Classical Mechanics and Thermodynamics

Kinematics: Position, Velocity, and Acceleration

Kinematics is the part of mechanics that describes how objects move. It focuses on motion itself, without first worrying about what causes the motion. In this lesson, you will learn how to describe one-dimensional motion using position, velocity, and acceleration.

You will also learn the difference between scalar and vector quantities, and how to calculate average and instantaneous rates of change. These ideas are the foundation for understanding motion in science.

1. Describing Position

Position tells where an object is located compared with a chosen reference point, often called the origin. In one-dimensional motion, we usually describe position along a straight line, such as a road or a hallway.

We use the symbol \(x\) for position. Position includes both a number and a direction relative to the origin, so it is a vector quantity.

For example, if a student stands 3 meters to the right of the classroom door, we can write the position as \(x = +3\,\text{m}\). If the student stands 2 meters to the left of the door, the position is \(x = -2\,\text{m}\).

The sign of position matters. A positive sign might mean right, east, or forward, while a negative sign might mean left, west, or backward. The meaning depends on how the coordinate system is chosen.

2. Scalar and Vector Quantities

In physics, some quantities have only size, while others have both size and direction.

• Scalar quantity: has magnitude only. Examples: distance, time, speed.
• Vector quantity: has magnitude and direction. Examples: position, displacement, velocity, acceleration.

This difference is very important in kinematics. Two objects can move with the same speed but in opposite directions, so their velocities are different.

For example:

• Speed of \(5\,\text{m/s}\) is a scalar.
• Velocity of \(+5\,\text{m/s}\) or \(-5\,\text{m/s}\) is a vector because direction matters.

3. Distance and Displacement

Students often confuse distance and displacement. They are not the same.

• Distance is the total length of the path traveled. It is a scalar.
• Displacement is the change in position. It is a vector.

Displacement is found by subtracting the initial position from the final position:

Here, \(\Delta x\) means “change in position,” \(x_f\) is final position, and \(x_i\) is initial position.

If a person walks 4 meters right and then 1 meter left, the distance traveled is \(5\,\text{m}\), but the displacement is \(+3\,\text{m}\).

4. Velocity: How Fast Position Changes

Velocity tells how quickly position changes and in what direction. Since direction matters, velocity is a vector.

The average velocity over a time interval is:

Here, \(\Delta t\) means the change in time.

Average velocity depends on displacement, not total distance. If an object returns to where it started, its displacement is zero, so its average velocity is zero, even if it traveled a long distance.

Instantaneous velocity means the velocity at one specific moment in time. A car’s speedometer gives information about its speed at that instant. If direction is included, that is instantaneous velocity.

In many 10th Grade problems, you may estimate instantaneous velocity from a graph or be told it directly. Conceptually, it means the rate of change of position at one exact moment.

5. Speed vs. Velocity

Speed and velocity are related, but they are not identical.

• Speed is how fast an object moves. It is a scalar.
• Velocity is speed with direction. It is a vector.

Average speed is:

Average velocity is:

These can give different values when direction changes during motion.

6. Acceleration: How Velocity Changes

Acceleration describes how velocity changes over time. Since velocity includes direction, acceleration can happen when speed changes, when direction changes, or both. In one-dimensional motion, we usually focus on speeding up, slowing down, or reversing direction along a straight line.

The formula for average acceleration is:

Here, \(v_f\) is final velocity and \(v_i\) is initial velocity.

Instantaneous acceleration is the acceleration at one exact moment. Like instantaneous velocity, it describes what is happening at a specific time.

Acceleration is a vector quantity. The sign tells direction:

• A positive acceleration points in the positive direction.
• A negative acceleration points in the negative direction.

Be careful: a negative acceleration does not always mean an object is slowing down. It depends on the direction of the velocity.

• If velocity and acceleration are in the same direction, the object speeds up.
• If velocity and acceleration are in opposite directions, the object slows down.

For example:

• \(v = +4\,\text{m/s}\), \(a = +2\,\text{m/s}^2\): speeding up.
• \(v = +4\,\text{m/s}\), \(a = -2\,\text{m/s}^2\): slowing down.
• \(v = -4\,\text{m/s}\), \(a = -2\,\text{m/s}^2\): speeding up in the negative direction.

7. Units in Kinematics

• Position and displacement: meters \((\text{m})\)
• Time: seconds \((\text{s})\)
• Velocity: meters per second \((\text{m/s})\)
• Acceleration: meters per second squared \((\text{m/s}^2)\)

Always include units in your answers. Units help show what a quantity means and can help you catch mistakes.

8. Motion Graphs

Graphs are useful for showing how motion changes over time. In one-dimensional kinematics, the two most common graphs are position-time and velocity-time graphs.

Position-Time Graph

• The vertical axis shows position.
• The horizontal axis shows time.
• The slope of the graph represents velocity.

What the slope means:

• Positive slope: positive velocity.
• Negative slope: negative velocity.
• Zero slope: object is not moving.
• Steeper slope: greater speed.

If the graph is curved, the velocity is changing, which means there is acceleration.

Velocity-Time Graph

• The vertical axis shows velocity.
• The horizontal axis shows time.
• The slope of the graph represents acceleration.

What the slope means:

• Positive slope: positive acceleration.
• Negative slope: negative acceleration.
• Zero slope: constant velocity.

Also, the area under a velocity-time graph represents displacement, but at this level the most important idea is that the slope tells acceleration.

9. Constant Velocity and Constant Acceleration

If an object has constant velocity, its velocity does not change. That means its acceleration is zero.

If an object has constant acceleration, its velocity changes by equal amounts in equal time intervals. A common example is an object in free fall near Earth, ignoring air resistance.

When acceleration is constant, one useful equation is:

This equation connects final velocity, initial velocity, acceleration, and time.

Another useful equation for position is:

In this lesson, the main goal is to understand what these quantities mean. As you solve more problems, you will become more comfortable choosing the correct equation.

10. Worked Examples

Example 1: Finding Displacement

A runner starts at \(x_i = 2\,\text{m}\) and finishes at \(x_f = 11\,\text{m}\). Find the displacement.

Step 1: Use the formula

Step 2: Substitute the values

Answer: The displacement is \(+9\,\text{m}\).

The positive sign means the runner ended 9 meters in the positive direction from where they started.

Example 2: Average Velocity

A toy car moves from \(x_i = -4\,\text{m}\) to \(x_f = 8\,\text{m}\) in \(6\,\text{s}\). Find the average velocity.

Step 1: Find displacement

Step 2: Use the average velocity formula

Step 3: Substitute the values

Answer: The average velocity is \(+2\,\text{m/s}\).

Example 3: Average Speed vs. Average Velocity

A student walks 10 meters east in 5 seconds, then 10 meters west in 5 seconds, returning to the starting point.

Find the average speed.

Total distance:

Total time:

Average speed:

Find the average velocity.

The student returns to the starting point, so displacement is:

Average velocity:

Answer:

• Average speed = \(2\,\text{m/s}\)
• Average velocity = \(0\,\text{m/s}\)

This example shows why speed and velocity are not the same.

Example 4: Average Acceleration

A bicycle moves at \(v_i = 3\,\text{m/s}\) and speeds up to \(v_f = 11\,\text{m/s}\) in \(4\,\text{s}\). Find the average acceleration.

Step 1: Use the formula

Step 2: Substitute the values

Answer: The average acceleration is \(+2\,\text{m/s}^2\).

This positive acceleration means the bicycle’s velocity is increasing in the positive direction.

11. Common Mistakes to Avoid

• Confusing distance with displacement.
• Confusing speed with velocity.
• Forgetting that vectors need direction or a sign.
• Ignoring negative signs in position, velocity, or acceleration.
• Using the wrong formula for average speed instead of average velocity.
• Forgetting units in the final answer.

12. How These Ideas Connect

These three quantities are closely connected:

• Position tells where the object is.
• Velocity tells how position changes over time.
• Acceleration tells how velocity changes over time.

You can think of them as a chain:

Change in position gives velocity, and change in velocity gives acceleration.

If position changes quickly, velocity is large. If velocity changes quickly, acceleration is large.

Brief Summary

Kinematics describes motion using position, velocity, and acceleration. Position and displacement describe where an object is and how its position changes. Velocity describes how fast position changes, while acceleration describes how fast velocity changes.

Remember that distance and speed are scalars, while displacement, velocity, and acceleration are vectors. In one-dimensional motion, signs such as positive and negative help show direction. Understanding these ideas will help you analyze motion clearly and solve many science problems.

Motion Graph Analysis helps us understand how an object moves by reading information from graphs. In 10th Grade science, the three most important motion graphs are position-time graphs, velocity-time graphs, and sometimes acceleration-time graphs. Each graph tells us something different about motion.

If you can read the slope of a graph and the area under a graph, you can learn a lot about an object's movement. In this lesson, you will learn how to find velocity from a position-time graph, acceleration from a velocity-time graph, and displacement from the area under a velocity-time graph.

First, remember these basic ideas:

• Position tells where an object is.
• Velocity tells how fast and in what direction the position is changing.
• Acceleration tells how fast the velocity is changing.
• Displacement is the change in position.

These ideas are connected by simple relationships:

$$\text{velocity} = \frac{\Delta \text{position}}{\Delta \text{time}}$$

$$\text{acceleration} = \frac{\Delta \text{velocity}}{\Delta \text{time}}$$

Here, the symbol \(\Delta\) means change in. For example, \(\Delta x\) means change in position, and \(\Delta t\) means change in time.

1. Position-Time Graphs

A position-time graph shows time on the horizontal axis and position on the vertical axis. The most important thing to study on this graph is the slope.

The slope of a position-time graph gives velocity:

$$v = \frac{\Delta x}{\Delta t}$$

This means:

• A steeper slope means a greater speed.
• A positive slope means the object is moving in the positive direction.
• A negative slope means the object is moving in the negative direction.
• A zero slope means the object is not moving.

If the graph is a straight line, the slope stays the same, so the velocity is constant. If the graph curves, then the velocity is changing.

2. Velocity-Time Graphs

A velocity-time graph shows time on the horizontal axis and velocity on the vertical axis. This graph gives two important pieces of information:

• The slope tells acceleration.
• The area under the graph tells displacement.

The slope of a velocity-time graph is:

$$a = \frac{\Delta v}{\Delta t}$$

This means:

• A positive slope means positive acceleration.
• A negative slope means negative acceleration.
• A zero slope means constant velocity and zero acceleration.

The area under the velocity-time graph gives displacement:

$$\text{displacement} = \text{area under the } v\text{-}t \text{ graph}$$

This works because velocity tells how position changes over time. When you multiply velocity by time, you get a change in position.

For simple shapes:

• Rectangle area: \(A = bh\)
• Triangle area: \(A = \frac{1}{2}bh\)

On a velocity-time graph, the base is usually time and the height is velocity.

3. Positive and Negative Areas

Displacement depends on direction. So on a velocity-time graph:

• Area above the time axis is positive displacement.
• Area below the time axis is negative displacement.

If an object moves forward and then backward, the positive and negative areas can cancel. That means the total displacement may be small even if the object traveled a large distance.

Distance and displacement are not always the same:

• Displacement includes direction.
• Distance is the total path traveled and is always positive.

4. How to Analyze a Motion Graph

When you see a motion graph, use these steps:

1. Read the axes carefully. Check what each axis represents and the units.
2. Look for straight sections, flat sections, or curved sections.
3. Find the slope if you need velocity or acceleration.
4. Find the area if you need displacement from a velocity-time graph.
5. Pay attention to positive and negative values.

Worked Example 1: Velocity from a Position-Time Graph

A car's position changes from \(2\text{ m}\) at \(t=1\text{ s}\) to \(14\text{ m}\) at \(t=4\text{ s}\). Find the velocity.

Step 1: Use the slope formula.

$$v = \frac{\Delta x}{\Delta t} = \frac{14 - 2}{4 - 1}$$

Step 2: Simplify.

$$v = \frac{12}{3} = 4\text{ m/s}$$

Answer: The car's velocity is \(4\text{ m/s}\).

This tells us the car moves in the positive direction at a constant speed if the graph is a straight line between those points.

Worked Example 2: Interpreting Slope on a Position-Time Graph

An object's position-time graph has three sections:

• From \(0\) to \(3\text{ s}\), the line slopes upward.
• From \(3\) to \(5\text{ s}\), the line is flat.
• From \(5\) to \(7\text{ s}\), the line slopes downward.

What is happening in each section?

Section 1: Upward slope means positive velocity. The object is moving forward.

Section 2: Flat line means zero slope, so velocity is zero. The object is stopped.

Section 3: Downward slope means negative velocity. The object is moving back toward the starting direction.

Answer: The object moves forward, stops, and then moves backward.

Worked Example 3: Acceleration from a Velocity-Time Graph

A runner's velocity changes from \(3\text{ m/s}\) at \(t=2\text{ s}\) to \(11\text{ m/s}\) at \(t=6\text{ s}\). Find the acceleration.

Step 1: Use the slope formula for a velocity-time graph.

$$a = \frac{\Delta v}{\Delta t} = \frac{11 - 3}{6 - 2}$$

Step 2: Simplify.

$$a = \frac{8}{4} = 2\text{ m/s}^2$$

Answer: The runner's acceleration is \(2\text{ m/s}^2\).

This means the velocity increases by \(2\text{ m/s}\) every second.

Worked Example 4: Displacement from the Area Under a Velocity-Time Graph

An object moves with a velocity of \(6\text{ m/s}\) for \(5\text{ s}\). Find the displacement.

On a velocity-time graph, this makes a rectangle with:

• base \(= 5\text{ s}\)
• height \(= 6\text{ m/s}\)

Step 1: Find the area.

$$A = bh = 5 \times 6 = 30$$

Step 2: Write the correct unit.

$$\text{displacement} = 30\text{ m}$$

Answer: The displacement is \(30\text{ m}\).

Now consider a slightly harder case. Suppose an object moves at \(8\text{ m/s}\) for \(4\text{ s}\), then at \(-4\text{ m/s}\) for \(2\text{ s}\).

First area:

$$A_1 = 8 \times 4 = 32\text{ m}$$

Second area is below the axis, so it is negative:

$$A_2 = (-4) \times 2 = -8\text{ m}$$

Total displacement:

$$\text{displacement} = 32 + (-8) = 24\text{ m}$$

Answer: The total displacement is \(24\text{ m}\).

5. Common Mistakes to Avoid

• Mixing up slope and area. On a position-time graph, slope gives velocity. On a velocity-time graph, slope gives acceleration and area gives displacement.
• Forgetting direction. Negative velocity or negative area means motion in the negative direction.
• Ignoring units. Velocity is usually in \(\text{m/s}\), acceleration in \(\text{m/s}^2\), and displacement in \(\text{m}\).
• Confusing displacement with distance. Displacement can be negative, but distance cannot.

6. Quick Comparison Chart

• Position-time graph: slope = velocity
• Velocity-time graph: slope = acceleration
• Velocity-time graph: area under curve = displacement

7. Final Idea

Motion graphs are like pictures of movement. Instead of only reading numbers, you can understand how an object moves over time. A rising line, a flat line, a steep slope, or a shaded area all tell a story about motion.

Once you know what slope and area mean, graph analysis becomes much easier. Always ask yourself: What kind of graph is this? Then decide whether you need the slope or the area.

Brief Summary

To analyze motion graphs, remember that the slope of a position-time graph gives velocity, the slope of a velocity-time graph gives acceleration, and the area under a velocity-time graph gives displacement. Positive and negative values matter because motion has direction. By reading slopes and areas carefully, you can describe how objects move and solve many motion problems.

Kinematic Equations are formulas that describe how objects move when their acceleration is constant. These equations help us connect position, displacement, velocity, acceleration, and time.

In 10th Grade science, kinematic equations are especially useful for studying motion in a straight line, such as a car speeding up, a bike slowing down, or an object falling under gravity.

To use these equations well, you need to understand the main motion quantities:

• Displacement \,  how far and in what direction an object moves
• Initial velocity \,  the starting speed and direction, written as \(v_i\)
• Final velocity \,  the ending speed and direction, written as \(v_f\)
• Acceleration \,  how quickly velocity changes, written as \(a\)
• Time \,  how long the motion lasts, written as \(t\)

These quantities are related by a set of equations called the kinematic equations.

Important condition: These equations only work when acceleration stays constant. If acceleration changes during the motion, these formulas do not apply directly.

Here are the main kinematic equations:

1. Final velocity after some time:

2. Displacement using average velocity:

3. Displacement using initial velocity:

4. Final velocity without time:

Each equation is useful in different situations. The best equation depends on which values you know and which value you are trying to find.

How to choose the right equation

• If you know time and want velocity, use \(v_f = v_i + at\).
• If you know time and want displacement, use \(d = v_i t + \frac{1}{2}at^2\).
• If you know both initial and final velocity and time, use \(d = \left(\frac{v_i + v_f}{2}\right)t\).
• If time is missing, use \(v_f^2 = v_i^2 + 2ad\).

Using signs correctly is very important in kinematics. Motion has direction, so positive and negative values matter.

You should first choose a positive direction. For example, you might choose rightward or upward as positive. Then:

• Velocities in the positive direction are positive.
• Velocities in the opposite direction are negative.
• Acceleration is positive if it points in the positive direction.
• Acceleration is negative if it points in the opposite direction.

For example, if upward is positive, then gravity is negative because it pulls downward.

Free-fall motion is one of the most common uses of kinematic equations. In free fall, the only significant force acting is gravity, so the acceleration is constant.

Near Earth, the acceleration due to gravity is approximately:

If upward is positive, then:

If downward is positive, then:

The sign depends on the direction you choose, but you must stay consistent throughout the problem.

Step-by-step strategy for solving kinematics problems

1. Read the problem carefully.
2. List the known values: \(v_i\), \(v_f\), \(a\), \(t\), and \(d\).
3. Choose a positive direction.
4. Pick the equation that includes the knowns and the unknown.
5. Substitute carefully, including signs and units.
6. Solve algebraically.
7. Check whether the answer is reasonable.

Worked Example 1: Finding final velocity

A car starts at \(10\,\text{m/s}\) and accelerates at \(2\,\text{m/s}^2\) for \(5\) seconds. Find its final velocity.

Step 1: Identify known values

• \(v_i = 10\,\text{m/s}\)
• \(a = 2\,\text{m/s}^2\)
• \(t = 5\,\text{s}\)

Step 2: Choose the equation

We want final velocity, and we know time, so use:

Step 3: Substitute values

Answer: The final velocity is \(20\,\text{m/s}\).

Worked Example 2: Finding displacement

A runner moves at an initial velocity of \(3\,\text{m/s}\) and accelerates at \(1.5\,\text{m/s}^2\) for \(4\) seconds. How far does the runner travel?

Known values

• \(v_i = 3\,\text{m/s}\)
• \(a = 1.5\,\text{m/s}^2\)
• \(t = 4\,\text{s}\)

We want displacement, so use:

Substitute:

Answer: The runner travels \(24\,\text{m}\).

Worked Example 3: Solving without time

A skateboarder moving at \(6\,\text{m/s}\) speeds up at \(2\,\text{m/s}^2\) over a distance of \(20\,\text{m}\). Find the final velocity.

Known values

• \(v_i = 6\,\text{m/s}\)
• \(a = 2\,\text{m/s}^2\)
• \(d = 20\,\text{m}\)

Time is not given, so use:

Substitute:

Take the square root:

Answer: The skateboarder’s final velocity is about \(10.8\,\text{m/s}\).

Worked Example 4: Free-fall problem

A ball is thrown straight upward at \(15\,\text{m/s}\). How high does it go before stopping for an instant at the top?

Choose a positive direction

Let upward be positive. Then:

• \(v_i = +15\,\text{m/s}\)
• \(v_f = 0\,\text{m/s}\) at the top
• \(a = -9.8\,\text{m/s}^2\)

We want displacement, and time is not given, so use:

Substitute:

Solve:

Answer: The ball rises about \(11.5\,\text{m}\).

Common mistakes to avoid

• Using the wrong sign for acceleration: Gravity is negative if upward is positive.
• Mixing up speed and velocity: Velocity includes direction.
• Using the wrong equation: Pick the one that matches the known and unknown values.
• Forgetting units: Always include units like meters, seconds, and meters per second.
• Ignoring constant acceleration: These equations only work if acceleration does not change.

Helpful tips

• Write down all known values before starting.
• Circle the unknown quantity.
• Check whether the answer makes physical sense.
• For free fall, decide your positive direction before plugging in numbers.
• If an answer for distance is negative, check your signs and what the question is really asking.

Quick review of the four equations

• $$v_f = v_i + at$$
• $$d = \left(\frac{v_i + v_f}{2}\right)t$$
• $$d = v_i t + \frac{1}{2}at^2$$
• $$v_f^2 = v_i^2 + 2ad$$

If you practice identifying knowns, choosing directions, and selecting the correct formula, kinematics problems become much easier to solve.

Summary

Kinematic equations describe motion with constant acceleration. They allow you to calculate displacement, velocity, acceleration, and time using relationships between these quantities.

These equations are especially useful for straight-line motion and free-fall problems. Success with kinematics depends on choosing the correct equation, keeping track of signs, and solving carefully step by step.

Projectile Motion is the motion of an object that is launched into the air and then moves under the pull of gravity. Common examples include a kicked soccer ball, a thrown basketball, or water sprayed from a hose.

The key idea is that projectile motion is a form of two-dimensional motion. This means the object moves in both the horizontal direction and the vertical direction at the same time.

To understand projectile motion, we separate the motion into two parts:

• Horizontal motion: motion side to side
• Vertical motion: motion up and down

These two parts happen together, but they can be studied independently. This is the most important idea in projectile motion.

Assumption: In most 10th Grade physics problems, we ignore air resistance. That means the only force acting on the projectile after it is launched is gravity.

Because gravity pulls downward, it affects the vertical motion, but it does not affect the horizontal motion.

So:

• The horizontal velocity stays constant.
• The vertical velocity changes because of gravity.

This is why the path of a projectile is curved.

1. Horizontal Motion

If air resistance is ignored, there is no horizontal force on the object. With no horizontal force, there is no horizontal acceleration.

So the horizontal velocity stays the same throughout the motion:

$$v_x = \text{constant}$$

The horizontal distance traveled can be found using:

$$x = v_x t$$

where:

• \(x\) = horizontal distance
• \(v_x\) = horizontal velocity
• \(t\) = time

2. Vertical Motion

The vertical motion is affected by gravity, which causes a constant downward acceleration. Near Earth, this acceleration is approximately:

$$g = 9.8\text{ m/s}^2$$

Since gravity acts downward, we usually write vertical acceleration as:

$$a_y = -9.8\text{ m/s}^2$$

The negative sign shows that the acceleration is downward if we choose upward as the positive direction.

The vertical motion follows the same equations used for any motion with constant acceleration:

$$v_y = v_{iy} + a_y t$$

$$y = v_{iy} t + \frac{1}{2}a_y t^2$$

$$v_y^2 = v_{iy}^2 + 2a_y y$$

where:

• \(v_y\) = vertical velocity at time \(t\)
• \(v_{iy}\) = initial vertical velocity
• \(y\) = vertical displacement
• \(a_y\) = vertical acceleration

3. Launching at an Angle

When an object is launched at an angle, its initial velocity must be split into horizontal and vertical parts. This process is called resolving components.

If the object is launched with initial speed \(v_i\) at angle \(\theta\):

$$v_{ix} = v_i\cos\theta$$

$$v_{iy} = v_i\sin\theta$$

These components tell us how fast the object starts moving horizontally and vertically.

Important facts:

• \(v_{ix}\) stays constant during the motion.
• \(v_{iy}\) changes because of gravity.
• At the highest point, the vertical velocity is \(0\).

4. Shape of the Path

A projectile follows a curved path called a parabola. It rises, slows down in the vertical direction, reaches a highest point, and then falls back down.

Even at the highest point, the object is still moving horizontally unless something stops it.

5. Time of Flight, Maximum Height, and Range

For projectiles that land at the same height from which they were launched, some useful ideas are:

• Time up equals time down.
• The vertical velocity on the way up and down is symmetric.
• The horizontal motion continues at constant speed for the entire flight.

Maximum height happens when \(v_y = 0\).

Using

$$v_y = v_{iy} + a_y t$$

we can find the time to reach the top:

$$0 = v_{iy} - 9.8t$$

$$t = \frac{v_{iy}}{9.8}$$

Then substitute that time into the vertical displacement formula to find the maximum height.

Range means the total horizontal distance traveled. Once we know the total time in the air, we can calculate range with:

$$\text{range} = v_x t$$

6. Special Case: Horizontal Launch

Sometimes an object is launched horizontally, like a ball rolling off a table. In this case:

• The initial vertical velocity is zero: \(v_{iy} = 0\)
• The horizontal velocity is still constant.
• The object begins falling immediately due to gravity.

This means horizontal launch problems are solved by combining:

• constant velocity motion horizontally, and
• free-fall motion vertically.

Worked Example 1: Horizontal Launch from a Table

A ball rolls off a table with a horizontal speed of \(4.0\text{ m/s}\). It takes \(0.50\text{ s}\) to hit the floor. How far from the table does it land?

Step 1: Identify the horizontal motion equation.

$$x = v_x t$$

Step 2: Substitute the known values.

$$x = (4.0)(0.50)$$

$$x = 2.0\text{ m}$$

Answer: The ball lands \(2.0\text{ m}\) from the table.

What to notice: We did not need gravity to find the horizontal distance once the time was known. Horizontal and vertical motion are connected only by time.

Worked Example 2: Vertical Motion in a Projectile

A ball is launched upward with an initial vertical velocity of \(14\text{ m/s}\). How long does it take to reach its highest point?

At the highest point:

$$v_y = 0$$

Use:

$$v_y = v_{iy} + a_y t$$

$$0 = 14 + (-9.8)t$$

$$0 = 14 - 9.8t$$

$$9.8t = 14$$

$$t = \frac{14}{9.8} \approx 1.43\text{ s}$$

Answer: It takes about \(1.43\text{ s}\) to reach the highest point.

Worked Example 3: Launch at an Angle

A projectile is launched at \(20\text{ m/s}\) at an angle of \(30^\circ\). Find its initial horizontal and vertical velocity components.

Use the component formulas:

$$v_{ix} = v_i\cos\theta$$

$$v_{iy} = v_i\sin\theta$$

Horizontal component:

$$v_{ix} = 20\cos 30^\circ$$

$$v_{ix} = 20(0.866) \approx 17.3\text{ m/s}$$

Vertical component:

$$v_{iy} = 20\sin 30^\circ$$

$$v_{iy} = 20(0.5) = 10\text{ m/s}$$

Answer:

• Initial horizontal velocity = \(17.3\text{ m/s}\)
• Initial vertical velocity = \(10\text{ m/s}\)

What to notice: The launch speed is split into two perpendicular directions. The horizontal part controls how fast it moves forward, and the vertical part controls how high it goes.

Worked Example 4: Finding Total Time and Range

A ball is launched at \(20\text{ m/s}\) at \(30^\circ\) and lands at the same height. Find:

• the total time in the air
• the horizontal range

From Example 3, the components are:

• \(v_{ix} = 17.3\text{ m/s}\)
• \(v_{iy} = 10\text{ m/s}\)

Step 1: Find time to reach the top.

$$t_{\text{up}} = \frac{v_{iy}}{9.8} = \frac{10}{9.8} \approx 1.02\text{ s}$$

Step 2: Find total time.

Since it lands at the same height:

$$t_{\text{total}} = 2(1.02) = 2.04\text{ s}$$

Step 3: Find horizontal range.

$$x = v_x t$$

$$x = 17.3 \times 2.04$$

$$x \approx 35.3\text{ m}$$

Answer:

• Total time in air = \(2.04\text{ s}\)
• Range = \(35.3\text{ m}\)

Common Mistakes to Avoid

• Mixing horizontal and vertical quantities. Use horizontal equations for side-to-side motion and vertical equations for up-and-down motion.
• Forgetting that gravity only affects vertical motion. Horizontal velocity stays constant if air resistance is ignored.
• Using the total speed as both components. When launched at an angle, split the velocity into \(x\)- and \(y\)-components.
• Forgetting signs. If upward is positive, then gravity is negative: \(a_y = -9.8\text{ m/s}^2\).
• Thinking the object stops at the top. Only the vertical velocity becomes zero there. The projectile still has horizontal motion.

Problem-Solving Steps

1. Draw or imagine the projectile path.
2. Separate the motion into horizontal and vertical parts.
3. List what is known for each direction.
4. If launched at an angle, find velocity components using sine and cosine.
5. Use vertical motion equations to find time, height, or vertical velocity.
6. Use the same time in the horizontal equation to find distance.

Quick Review of Main Equations

Horizontal:

$$x = v_x t$$

Vertical:

$$v_y = v_{iy} + a_y t$$

$$y = v_{iy} t + \frac{1}{2}a_y t^2$$

$$v_y^2 = v_{iy}^2 + 2a_y y$$

Components of launch velocity:

$$v_{ix} = v_i\cos\theta$$

$$v_{iy} = v_i\sin\theta$$

Brief Summary

Projectile motion is the motion of an object moving through the air under the influence of gravity. The most important idea is that the horizontal and vertical motions are independent.

Horizontally, the object moves with constant velocity. Vertically, it accelerates downward at \(9.8\text{ m/s}^2\). By splitting the motion into these two directions, you can solve for time, height, speed, and range.

Newton's First Law and Inertial Reference Frames

When we watch objects move, it may seem like motion always needs a cause. For example, a rolling ball slows down, and a car stops if the driver does not press the gas pedal. Because of everyday experience, people often think that a force is needed to keep an object moving.

Newton's First Law shows that this idea is not quite correct. An object does not need a force to keep moving at a constant speed in a straight line. Instead, a force is needed to change its motion.

This lesson explains Newton's First Law, the idea of inertia, and what scientists mean by an inertial reference frame. These ideas are important because they help us describe motion correctly and understand when Newton's laws can be used directly.

1. Newton's First Law

Newton's First Law states:

An object at rest stays at rest, and an object in motion stays in motion with constant velocity unless acted on by a net external force.

There are two key parts of this law:

• At rest stays at rest: A book on a table will remain there unless something pushes, pulls, or otherwise acts on it.
• In motion stays in motion: A moving object will keep going in a straight line at constant speed unless a net external force changes its motion.

The phrase constant velocity is very important. Velocity includes both speed and direction. So if either speed changes or direction changes, then velocity changes.

2. Net External Force

A force is a push or a pull. Sometimes more than one force acts on an object at the same time. The net force is the overall force after combining all the forces.

If the net force is zero, then the object's motion does not change. This means:

• an object at rest remains at rest, or
• an object already moving continues with constant velocity.

We often write this idea as:

$$F_{\text{net}} = 0 \Rightarrow \text{no change in velocity}$$

This does not mean the object has no forces acting on it. It means the forces are balanced so that together they add to zero.

For example, a book resting on a table has:

• gravity pulling downward, and
• the table pushing upward.

These forces balance, so the net force is zero. The book does not accelerate.

3. Inertia

The tendency of an object to resist changes in its motion is called inertia. Inertia is why objects do not easily start moving, stop moving, or change direction.

Mass is the measure of inertia. This means:

• an object with more mass has more inertia,
• an object with less mass has less inertia.

A heavy shopping cart is harder to start moving than an empty cart. It is also harder to stop once it is moving. This is because the heavier cart has more mass, so it has more inertia.

In simple terms, mass tells us how strongly an object resists a change in motion.

4. Everyday Examples of Newton's First Law

• Seatbelts in a car: When a car stops suddenly, your body tends to keep moving forward because of inertia. The seatbelt provides the force that stops you safely.
• Tablecloth trick: If a tablecloth is pulled quickly from under dishes, the dishes tend to remain nearly at rest because of inertia.
• A hockey puck on ice: It keeps sliding much longer than a ball rolling on the ground because there is less friction.
• Riding a bus: When the bus starts moving, your body seems to lean backward because your body resists the change from rest to motion.

In many real-life situations, friction and air resistance act as forces that slow objects down. That is why moving objects in everyday life often do not keep moving forever.

5. Why Objects Seem to Stop on Their Own

If Newton's First Law says moving objects keep moving, why does a soccer ball eventually stop rolling?

The answer is that a net external force is acting on it. Forces such as friction with the ground and air resistance oppose the ball's motion. These forces change the velocity, so the ball slows down and stops.

If there were no friction or air resistance, the ball would continue moving much longer. In an ideal situation with zero net force, it would keep moving at constant velocity.

6. Inertial Reference Frames

A reference frame is a point of view from which motion is observed and measured. For example, you might describe motion from the ground, from inside a moving train, or from inside a car.

An inertial reference frame is a reference frame in which Newton's First Law works directly. In this kind of frame:

• if the net external force on an object is zero, the object stays at rest or moves with constant velocity.

Usually, a reference frame that is not accelerating is treated as inertial. A classroom, a road, or the ground is often close enough to an inertial frame for everyday problems.

7. Non-Inertial Reference Frames

A non-inertial reference frame is a frame that is accelerating. This includes speeding up, slowing down, or turning.

Inside a car that suddenly brakes, passengers seem to lurch forward. Inside the car, it may feel like some invisible force is pushing them. But really, their bodies are continuing in their previous state of motion because of inertia.

This is why motion can seem strange in an accelerating frame. Newton's First Law does not appear to work in the usual way unless we account for the acceleration of the frame.

8. Comparing Inertial and Non-Inertial Frames

• Inertial frame: A train moving at constant speed in a straight line. If you toss a ball straight up, it comes back down to your hand normally.
• Non-inertial frame: A train that is speeding up. If you toss a ball straight up, its motion may seem shifted because the train is accelerating.

So, when using Newton's laws, we must be careful about the frame from which we observe the motion.

9. Connection Between Mass and Resistance to Motion Changes

The concept statement says that mass dictates an object's resistance to changes in its state of motion in the absence of a net external force. This means mass controls how much inertia an object has.

Imagine pushing two boxes across a smooth floor:

• one box has a mass of 5 kg,
• the other has a mass of 20 kg.

The 20 kg box resists changes in motion more strongly. It is harder to start moving, harder to stop, and harder to change direction. This is because greater mass means greater inertia.

10. Worked Examples

Example 1: Book on a table

A book sits motionless on a table. Is Newton's First Law satisfied?

Step 1: Identify the forces.

• Gravity pulls the book downward.
• The table pushes upward on the book.

Step 2: Compare the forces.

These forces are equal in size and opposite in direction, so the net force is zero.

$$F_{\text{net}} = 0$$

Step 3: Decide what happens.

Because the net force is zero, the book remains at rest. This matches Newton's First Law.

Answer: Yes. The book stays at rest because the forces are balanced.

Example 2: Sliding puck

A hockey puck slides across smooth ice. After being hit, it moves in a straight line at constant speed for a while. What does Newton's First Law say about this motion?

Step 1: Look at the motion.

The puck's velocity is not changing. It is moving at constant speed in a straight line.

Step 2: Apply Newton's First Law.

If velocity stays constant, then the net external force must be zero or very close to zero.

$$F_{\text{net}} \approx 0$$

Step 3: Explain why this is possible.

The ice provides very little friction, so there is little force opposing the motion.

Answer: The puck continues moving because there is little to no net force changing its motion.

Example 3: Comparing inertia

A 2 kg ball and a 10 kg ball are both at rest. Which one is harder to start moving, and why?

Step 1: Recall the meaning of inertia.

Inertia is resistance to a change in motion.

Step 2: Relate inertia to mass.

More mass means more inertia.

Step 3: Compare the objects.

The 10 kg ball has more mass than the 2 kg ball.

Answer: The 10 kg ball is harder to start moving because it has greater mass and therefore greater inertia.

Example 4: Car braking and reference frames

A passenger is riding in a car. The car suddenly brakes, and the passenger seems to move forward. Does this mean a forward force pushed the passenger?

Step 1: Think about inertia.

Before braking, the passenger and car were moving forward together.

Step 2: What happens when the car slows?

The car slows because braking provides a backward force on the car. The passenger's body tends to keep moving forward at the original velocity.

Step 3: Interpret the observation.

From inside the braking car, it may seem like the passenger is pushed forward. But in an inertial frame outside the car, the passenger is simply continuing in motion because of inertia.

Answer: No. The passenger moves forward relative to the car because the car slows down, while the passenger's body resists the change in motion.

11. Common Mistakes to Avoid

• Mistake 1: Thinking a force is needed to keep something moving.
  Actually, a force is needed to change motion, not maintain constant velocity.
• Mistake 2: Thinking zero motion means zero forces.
  An object can be at rest with balanced forces.
• Mistake 3: Confusing speed with velocity.
  Velocity includes direction, so turning is also a change in velocity.
• Mistake 4: Forgetting the role of the reference frame.
  Newton's First Law is applied most directly in inertial reference frames.

12. Key Ideas to Remember

• Newton's First Law is also called the law of inertia.
• If the net external force is zero, velocity does not change.
• Inertia is an object's resistance to changes in motion.
• Mass measures inertia.
• More mass means greater resistance to starting, stopping, or changing direction.
• An inertial reference frame is a frame where Newton's First Law works normally.
• An accelerating frame is non-inertial.

Brief Summary

Newton's First Law says that objects keep their current state of motion unless a net external force changes it. This means an object at rest stays at rest, and an object in motion continues with constant velocity. The reason is inertia, which is measured by mass. In inertial reference frames, these ideas describe motion clearly and help us understand why balanced forces do not change motion, while unbalanced forces do.

Newton's Second Law and Net Force

When an object speeds up, slows down, or changes direction, something causes that change in motion. In physics, that cause is force. Newton's Second Law explains how the forces acting on an object are related to its mass and its acceleration.

This lesson will help you understand how to find net force, draw free-body diagrams, and use the equation \(F=ma\) to calculate an object's acceleration in situations with more than one force.

1. What is Newton's Second Law?

Newton's Second Law says that the net force acting on an object equals the object's mass times its acceleration.

In this equation:

• \(F_{\text{net}}\) is the net force, measured in newtons (N)
• \(m\) is the mass, measured in kilograms (kg)
• \(a\) is the acceleration, measured in meters per second squared \((m/s^2)\)

This law tells us two important ideas:

• If the net force increases, the acceleration increases.
• If the mass increases, the acceleration decreases for the same net force.

So, a lighter object is easier to accelerate than a heavier object if the same force is applied.

2. What is net force?

Net force is the overall force acting on an object after all the individual forces are combined. Because force has both size and direction, you must include direction when adding forces.

If forces act in the same direction, add them. If they act in opposite directions, subtract them.

For example:

• If \(10\,N\) acts to the right and \(4\,N\) acts to the right, the net force is \(14\,N\) to the right.
• If \(10\,N\) acts to the right and \(4\,N\) acts to the left, the net force is \(6\,N\) to the right.

If the net force is zero, then the object does not accelerate. It may stay at rest or continue moving at constant velocity.

3. Common forces you may see

In many 10th Grade physics problems, you will work with a small set of common forces:

• Weight: the force of gravity pulling downward
• Normal force: the support force from a surface, usually upward
• Applied force: a push or pull from a person or object
• Friction: a force that opposes motion between surfaces
• Tension: a pulling force in a rope or string
• Air resistance: a force from air that opposes motion

The weight of an object can be found using

where \(g \approx 9.8\,m/s^2\) on Earth. In many class problems, this may be rounded to \(10\,m/s^2\).

4. Free-body diagrams

A free-body diagram is a simple drawing that shows all the forces acting on one object. It helps organize the problem before doing any math.

To draw a free-body diagram:

1. Draw the object as a box or dot.
2. Draw arrows for each force acting on the object.
3. Label each arrow with the force name or value.
4. Make the arrow point in the direction of the force.

Example of a book resting on a table:

• Weight acts downward.
• Normal force acts upward.

If the book is not accelerating up or down, those two forces are equal in size, so the net force is zero.

5. Net force in different directions

Forces often act in both horizontal and vertical directions. A very useful idea is to treat each direction separately.

That means you can think about:

• Horizontal forces left and right
• Vertical forces up and down

If an object moves across a floor without moving up or down, then the vertical forces usually balance. The acceleration comes from the horizontal net force.

For example, if a box is pushed across a flat floor:

• Applied force may act to the right
• Friction may act to the left
• Normal force acts up
• Weight acts down

If the box does not rise or sink, then the upward and downward forces cancel. You only need to find the net horizontal force to calculate the horizontal acceleration.

6. Using signs and direction

It is helpful to choose a positive direction. For example, you might choose right as positive and left as negative.

Then a force to the right is positive, and a force to the left is negative. This helps you combine forces correctly.

For horizontal motion:

Then use Newton's Second Law:

If your answer for acceleration is negative, that means the acceleration is in the negative direction you chose.

7. Resolved force vectors

Sometimes a force may not point exactly left, right, up, or down. In those cases, we resolve the force into horizontal and vertical parts. This means we split one angled force into two components.

At this level, the key idea is simple: one diagonal force can act like a combination of one horizontal force and one vertical force.

After splitting the force into components, add forces in each direction separately:

• Find net force in the \(x\)-direction
• Find net force in the \(y\)-direction

Then apply \(F=ma\) in each direction if needed.

8. Worked Example 1: One force on an object

A \(5\,kg\) cart is pushed with a net force of \(20\,N\) to the right. Find its acceleration.

Step 1: Write the equation.

Step 2: Solve for acceleration.

Step 3: Substitute values.

Answer: The cart accelerates at \(4\,m/s^2\) to the right.

This example shows that when you already know the net force, finding acceleration is direct.

9. Worked Example 2: Two opposite horizontal forces

A \(10\,kg\) box is pushed with \(35\,N\) to the right. Friction acts with \(15\,N\) to the left. Find the net force and acceleration.

Step 1: Find net force.

So the net force is \(20\,N\) to the right.

Step 2: Use Newton's Second Law.

Answer:

• Net force = \(20\,N\) to the right
• Acceleration = \(2\,m/s^2\) to the right

This example shows that you must combine all forces first before using \(F=ma\).

10. Worked Example 3: Horizontal and vertical forces

A \(4\,kg\) box is pulled across a flat floor. The applied force is \(18\,N\) to the right, and friction is \(6\,N\) to the left. The box does not move up or down.

Find the horizontal acceleration.

Step 1: Draw or imagine the free-body diagram.

• Applied force: right
• Friction: left
• Normal force: up
• Weight: down

Step 2: Look at vertical forces.

Because the box does not move up or down, the vertical forces balance. So the vertical net force is zero.

Step 3: Find horizontal net force.

Step 4: Find acceleration.

Answer: The box accelerates at \(3\,m/s^2\) to the right.

This example is important because many problems include balanced forces in one direction and unbalanced forces in another.

11. Worked Example 4: Weight and vertical acceleration

A \(2\,kg\) object is lifted upward by a rope with a tension force of \(25\,N\). Find its acceleration.

Step 1: Find the weight.

The weight acts downward with \(19.6\,N\).

Step 2: Find the net force.

Tension is upward, weight is downward.

So the net force is \(5.4\,N\) upward.

Step 3: Use Newton's Second Law.

Answer: The object accelerates upward at \(2.7\,m/s^2\).

This example shows that gravity must be included whenever vertical motion is involved.

12. A step-by-step strategy for solving problems

Use this method whenever you solve a Newton's Second Law problem:

1. Identify the object you are studying.
2. Draw a free-body diagram.
3. List all forces acting on the object.
4. Separate forces by direction: horizontal and vertical.
5. Find the net force in the direction of motion or acceleration.
6. Use $$a = \frac{F_{\text{net}}}{m}$$
7. State the direction of the acceleration.

13. Common mistakes to avoid

• Using one force instead of net force: Always combine all forces first.
• Ignoring direction: Right and left, or up and down, must be treated carefully.
• Forgetting gravity: If the problem involves vertical forces, include weight.
• Mixing balanced and unbalanced forces: Forces can cancel in one direction but not another.
• Leaving out units: Use newtons, kilograms, and \(m/s^2\).

14. Key ideas to remember

• Newton's Second Law is $$F_{\text{net}} = ma$$
• Acceleration is caused by net force, not just any single force.
• Net force includes both size and direction.
• Free-body diagrams help you organize forces clearly.
• In multi-force problems, add and subtract forces carefully by direction.
• If net force is zero, acceleration is zero.

Brief Summary

Newton's Second Law connects force, mass, and acceleration through the equation $$F_{\text{net}} = ma$$. To use it correctly, you must first find the net force by combining all forces acting on the object. Free-body diagrams make this easier by showing every force and its direction. Once the net force is known, divide by mass to find the acceleration and include the correct direction in your answer.

Newton's Third Law explains what happens when two objects interact. It is often stated as: For every action force, there is an equal and opposite reaction force.

This means that if object A pushes or pulls on object B, then object B pushes or pulls on object A with a force of the same size but in the opposite direction.

A very important idea is that these two forces act on different objects. Because they act on different objects, they do not cancel each other out.

Newton's Third Law helps explain many everyday situations, such as walking, swimming, jumping, rockets launching, and even sitting in a chair.

Newton's Third Law in symbols:

This says the force of A on B is equal in magnitude and opposite in direction to the force of B on A.

Forces that follow this rule are called interaction pairs or action-reaction pairs.

Main idea: Forces always come in pairs when two objects interact.

How to recognize an interaction pair

• The two forces are the same type of force. For example, both are contact forces, or both are gravitational forces.
• The two forces act on two different objects.
• The forces are equal in size.
• The forces act in opposite directions.
• The forces happen at the same time.

A common mistake is to choose two forces acting on the same object and call them an action-reaction pair. That is incorrect.

For example, if a book rests on a table, the table pushes up on the book, and Earth pulls down on the book with gravity. These two forces act on the same object, the book. So they are not an action-reaction pair.

Instead, the correct interaction pairs are:

• The table pushes up on the book, and the book pushes down on the table.
• The Earth pulls down on the book, and the book pulls up on the Earth.

Why action-reaction forces do not cancel

Students often wonder: if the forces are equal and opposite, why does anything move?

The answer is that the two forces act on different objects. Cancellation only happens when opposite forces act on the same object.

For motion, we look at the net force on one object at a time. If the forces on that one object are balanced, it will not speed up or slow down. If the forces on that one object are unbalanced, it will accelerate.

Example: standing on the ground

When you stand still, your feet push down on the ground. The ground pushes up on your feet.

These are an interaction pair:

• Feet push on ground.
• Ground pushes on feet.

At the same time, Earth pulls you downward with gravity. If you are standing still, the upward force from the ground and the downward gravitational force on you are balanced.

Notice the difference:

• Ground on you and gravity on you are forces on the same object, so we compare them for balance.
• You on ground and ground on you are forces on different objects, so they form an interaction pair.

Types of forces that can form interaction pairs

• Contact force: pushing, pulling, normal force, friction, tension
• Gravitational force: attraction between masses
• Electric or magnetic forces: if studied in class

No matter what type of force it is, if two objects interact, each object exerts a force on the other.

Worked Example 1: Book on a table

A book rests on a table. Identify one action-reaction pair.

Step 1: Look for two interacting objects. The book and the table are touching.

Step 2: Describe the force of one object on the other. The table pushes up on the book.

Step 3: Reverse the objects. The book pushes down on the table.

So the interaction pair is:

• Force of the table on the book
• Force of the book on the table

These forces are equal in size and opposite in direction.

Important: The table pushing up on the book and Earth pulling down on the book are not an interaction pair, because both forces act on the book.

Worked Example 2: Walking forward

When a person walks, why do they move forward?

Step 1: Identify the interaction. The foot pushes backward on the ground.

Step 2: Apply Newton's Third Law. The ground pushes forward on the foot with an equal and opposite force.

Step 3: Decide which force helps the person move. The forward force of the ground on the foot acts on the person, so it can make the person move forward.

This is why walking is possible. You push backward on the ground, and the ground pushes you forward.

Worked Example 3: Two skaters pushing apart

Two skaters stand still on smooth ice. Skater A pushes skater B. Both skaters move apart.

Question: If skater A pushes on skater B with a force of \(80\text{ N}\), how much force does skater B push on skater A?

Solution: By Newton's Third Law, the forces are equal in magnitude and opposite in direction.

So if skater A pushes skater B with \(80\text{ N}\), then skater B pushes skater A with \(80\text{ N}\) in the opposite direction.

Answer: \(80\text{ N}\), opposite in direction.

Why might they move differently? Even though the forces are equal, the skaters may have different masses. A smaller mass can have a larger acceleration.

Worked Example 4: Rocket launch

A rocket engine pushes hot gases downward. Explain how this makes the rocket move upward.

Step 1: Identify the first force. The rocket pushes the gases downward.

Step 2: Apply Newton's Third Law. The gases push the rocket upward with an equal and opposite force.

Step 3: Connect to motion. The upward force of the gases on the rocket can cause the rocket to accelerate upward if it is greater than the downward pull of gravity.

This is an important example because it shows that motion does not require pushing against the ground. The rocket moves because of the interaction between the rocket and the gases.

How to tell whether forces are an interaction pair or just balanced forces

1. Ask: What object is each force acting on?
2. If both forces act on the same object, they are not an action-reaction pair.
3. If the forces act on different objects and are equal and opposite, they can be an interaction pair.
4. Check that they come from the same interaction.

Example of balanced forces, not an interaction pair:

• A box sits on the floor.
• The floor pushes up on the box.
• Earth pulls down on the box.

These two forces may be equal, so the box stays still. But they are not an action-reaction pair because both act on the box.

Example of an interaction pair:

• The floor pushes up on the box.
• The box pushes down on the floor.

Important ideas to remember

• Action-reaction forces are always equal and opposite.
• They always act on different objects.
• They do not cancel each other because they are on different objects.
• To study motion, focus on all the forces acting on one object at a time.

Quick check questions

1. A swimmer pushes water backward. What is the reaction force?

Answer: The water pushes the swimmer forward.

2. A wall pushes back on your hand when you press on it. What is the action force?

Answer: Your hand pushes on the wall.

3. A ball falls toward Earth because Earth pulls on the ball. What is the reaction force?

Answer: The ball pulls on Earth.

Brief Summary

Newton's Third Law says that whenever two objects interact, they exert equal and opposite forces on each other. These forces are called interaction pairs. The most important thing to remember is that the two forces act on different objects, so they do not cancel. To decide whether two forces are an action-reaction pair, always check which objects the forces act on.

Friction and Drag are forces that oppose motion. They are very important in everyday life. Friction helps you walk without slipping, but it also makes bike tires wear down. Drag slows down cars, runners, and falling objects moving through air or water.

In this lesson, you will learn how to calculate static friction and kinetic friction using coefficients and normal force. You will also learn how air resistance, or drag, affects moving objects and how it can lead to terminal velocity.

Understanding these ideas helps explain why objects do not always keep speeding up, even when gravity or another force is acting. Instead, opposing forces can balance the motion.

1. What is friction?

Friction is a force that resists motion between two surfaces that are touching. It acts parallel to the surfaces and in the direction opposite the motion, or opposite the attempted motion.

For example, if you push a box to the right across the floor, friction acts to the left. If the box is not moving yet, friction still acts to oppose the push.

There are two main types of friction you need to know:

• Static friction: friction that acts when an object is not moving yet.
• Kinetic friction: friction that acts when an object is sliding.

2. The normal force

To calculate friction, we need the normal force. The normal force is the support force from a surface. It acts perpendicular to the surface.

On a flat horizontal surface, if there are no other vertical forces, the normal force equals the object's weight:

where:

• \(F_N\) = normal force in newtons (N)
• \(m\) = mass in kilograms (kg)
• \(g\) = gravitational field strength, about \(9.8\,\text{m/s}^2\) on Earth

This means heavier objects usually have a larger normal force, which often leads to more friction.

3. Static friction

Static friction keeps an object from starting to move. It adjusts its size to match the applied force, up to a maximum value.

The maximum static friction is:

where:

• \(f_s^{\text{max}}\) = maximum static friction
• \(\mu_s\) = coefficient of static friction
• \(F_N\) = normal force

The coefficient of friction is a number that describes how rough or sticky two surfaces are. A larger coefficient means more friction.

Important idea: static friction is not always equal to \(\mu_s F_N\). That formula gives the largest possible static friction. The actual static friction can be smaller if less force is needed to keep the object at rest.

So, for an object that is not moving:

If the applied force becomes greater than the maximum static friction, the object starts moving.

4. Kinetic friction

Once the object is sliding, the friction becomes kinetic friction. Kinetic friction usually stays close to a constant value.

The formula is:

where:

• \(f_k\) = kinetic friction
• \(\mu_k\) = coefficient of kinetic friction
• \(F_N\) = normal force

Usually, \(\mu_s > \mu_k\). That means it is often harder to start moving an object than to keep it moving.

5. Direction of friction

Friction always acts in the direction opposite the motion or opposite the tendency to move.

• If a box slides right, kinetic friction acts left.
• If you try to push a box right but it stays still, static friction acts left.

This matters when finding the net force, which determines acceleration.

Remember Newton's second law:

If friction is the only force opposing motion on a horizontal surface, then:

If the net force is positive, the object accelerates in that direction. If the net force is zero, the object's motion does not change.

6. Worked Example 1: Maximum static friction

A \(12\,\text{kg}\) box rests on a floor. The coefficient of static friction is \(\mu_s = 0.40\). Find the maximum static friction.

Step 1: Find the normal force.

Step 2: Use the static friction formula.

Answer: The maximum static friction is about \(47\,\text{N}\).

This means if you push with less than \(47\,\text{N}\), the box can stay at rest. If you push harder than that, the box will begin to move.

7. Worked Example 2: Will the object move?

A student pushes the same \(12\,\text{kg}\) box with a force of \(30\,\text{N}\). Will it move?

From Example 1, the maximum static friction is \(47\,\text{N}\).

The applied force is only \(30\,\text{N}\), which is less than \(47\,\text{N}\). Static friction will match the push and have a size of \(30\,\text{N}\) in the opposite direction.

So:

Answer: The box does not move.

This example shows why static friction changes as needed, up to its maximum value.

8. Worked Example 3: Kinetic friction and acceleration

A \(10\,\text{kg}\) sled is pulled across snow with a horizontal force of \(35\,\text{N}\). The coefficient of kinetic friction is \(\mu_k = 0.20\). Find the kinetic friction and the acceleration.

Step 1: Find the normal force.

Because the sled is on a horizontal surface:

Step 2: Find kinetic friction.

Step 3: Find the net force.

Step 4: Use Newton's second law.

Answer:

• Kinetic friction = \(19.6\,\text{N}\)
• Acceleration = \(1.54\,\text{m/s}^2\)

9. What is drag?

Drag is a friction-like force that acts on an object moving through a fluid. A fluid can be a liquid, like water, or a gas, like air.

Air resistance is a common type of drag. It acts opposite the direction of motion. If a skydiver is falling downward, drag acts upward.

Drag depends on several things:

• the speed of the object
• the shape of the object
• the size of the object
• the fluid it is moving through

A flat sheet of paper has more drag than a crumpled paper ball because its shape and area make it push against more air.

10. How drag changes with speed

One important difference between surface friction and drag is that drag often increases as speed increases.

At low speeds, drag may be modeled as roughly proportional to speed. At higher speeds, drag often increases much faster. In many 10th Grade problems, you only need the idea that faster motion causes greater drag.

This is why a cyclist feels much more air resistance when riding fast than when riding slowly.

11. Falling objects and terminal velocity

When an object falls, gravity pulls it downward. At the same time, air resistance pushes upward.

At first, when the object starts falling, its speed is small, so drag is small. Gravity is greater than drag, so the object accelerates downward.

As the object speeds up, drag becomes larger. This reduces the net force.

Eventually, drag becomes equal to the weight of the object. Then the forces balance:

So the net force is zero:

That means the acceleration becomes zero. The object continues falling, but now at a constant speed called terminal velocity.

Terminal velocity is the constant speed a falling object reaches when drag equals its weight.

12. Understanding terminal velocity

Terminal velocity does not mean the object stops moving. It means the object stops speeding up.

This is an important idea. Students sometimes confuse zero acceleration with zero velocity. At terminal velocity, acceleration is zero, but velocity is not zero.

The object is still moving downward, just at a constant speed.

13. Worked Example 4: Terminal velocity

A falling object has a weight of \(196\,\text{N}\). At one moment, the upward drag force is \(150\,\text{N}\).

Part A: Is the object still accelerating?

Find the net force downward:

Because the net force is downward, the object is still accelerating downward.

Part B: What happens when drag reaches \(196\,\text{N}\)?

Then:

Now the acceleration is zero, so the object has reached terminal velocity.

Answer: The object is still speeding up when drag is \(150\,\text{N}\). It reaches terminal velocity when drag becomes equal to its weight, \(196\,\text{N}\).

14. Comparing friction and drag

• Friction usually acts between solid surfaces that touch.
• Drag acts when an object moves through a fluid, such as air or water.
• Static friction prevents motion from starting.
• Kinetic friction opposes sliding motion.
• Drag usually increases as speed increases.

15. Common mistakes to avoid

• Mixing up static and kinetic friction: use static friction when the object is not moving, kinetic friction when it is sliding.
• Assuming static friction always equals \(\mu_s F_N\): that is only the maximum value.
• Forgetting the normal force: friction depends on \(F_N\), not directly on mass alone.
• Using the wrong direction: friction and drag always oppose motion or attempted motion.
• Thinking terminal velocity means stopping: it means moving at constant speed.

16. Problem-solving steps

When solving friction or drag problems, use this method:

1. Identify the forces acting on the object.
2. Find the normal force if needed.
3. Choose the correct friction formula:
   • Static: \(f_s \leq \mu_s F_N\)
   • Kinetic: \(f_k = \mu_k F_N\)
4. Find the net force.
5. Use \(F_{\text{net}} = ma\) if you need acceleration.
6. For falling objects, compare weight and drag to determine whether the object is accelerating or at terminal velocity.

Brief Summary

Friction is a force that opposes motion between surfaces in contact. Static friction prevents motion from starting and can vary up to a maximum of \(\mu_s F_N\). Kinetic friction acts during sliding and is given by \(f_k = \mu_k F_N\).

Drag is a force that opposes motion through air or water. As speed increases, drag usually increases. A falling object reaches terminal velocity when the upward drag equals the downward weight, making the net force zero and the speed constant.

Uniform Circular Motion happens when an object moves in a circle at a constant speed. Even though the speed stays the same, the object is still accelerating because its direction is always changing.

This idea is very important in mechanics because any object moving in a circle must have a net force toward the center of the circle. This inward force is called the centripetal force. The word centripetal means “center-seeking.”

In this lesson, you will learn how to describe uniform circular motion, calculate centripetal acceleration, and identify which real forces—such as tension, gravity, or friction—provide the needed inward force.

1. What makes circular motion different?

When an object moves in a straight line at constant speed, its velocity stays constant because both its speed and direction stay the same. In circular motion, the speed may stay constant, but the direction keeps changing every moment.

Since velocity includes direction, a change in direction means a change in velocity. A change in velocity means acceleration. So an object moving in a circle is accelerating even if its speed does not change.

That acceleration points toward the center of the circle. It is called centripetal acceleration.

2. Centripetal acceleration

The formula for centripetal acceleration is

$$a_c = \frac{v^2}{r}$$

where:

• \(a_c\) = centripetal acceleration in meters per second squared, \(\text{m/s}^2\)
• \(v\) = speed in meters per second, \(\text{m/s}\)
• \(r\) = radius of the circle in meters, \(\text{m}\)

This formula shows two important ideas:

• If speed increases, centripetal acceleration increases a lot because speed is squared.
• If the radius gets larger, centripetal acceleration gets smaller.

So a sharp turn at high speed needs a very large inward acceleration.

3. Centripetal force

Newton’s second law says that net force equals mass times acceleration:

$$F_{net} = ma$$

For circular motion, the net inward force is the centripetal force. So:

$$F_c = ma_c$$

Substituting the formula for centripetal acceleration gives:

$$F_c = m\frac{v^2}{r}$$

where:

• \(F_c\) = centripetal force in newtons, \(\text{N}\)
• \(m\) = mass in kilograms, \(\text{kg}\)
• \(v\) = speed in meters per second
• \(r\) = radius in meters

Important: centripetal force is not a new kind of force. It is the name for the net force directed toward the center. Different real forces can act as the centripetal force depending on the situation.

4. Real forces that can provide centripetal force

The centripetal force can come from different sources:

• Tension in a rope or string
• Friction between tires and the road
• Gravity between planets, moons, or satellites
• Normal force from a surface in some cases

When solving problems, do not write “centripetal force” as a separate extra force in a force diagram. Instead, identify the actual force that points inward.

For example:

• A ball tied to a string and swung in a circle: tension provides the centripetal force.
• A car turning on a flat road: friction provides the centripetal force.
• The Moon orbiting Earth: gravity provides the centripetal force.

5. Direction of velocity, acceleration, and force

In uniform circular motion:

• The velocity of the object is always tangent to the circle, meaning it points in the direction the object is moving.
• The centripetal acceleration points toward the center.
• The centripetal net force also points toward the center.

This means the velocity and centripetal acceleration point in different directions.

If the inward force suddenly disappears, the object does not keep moving in a circle. It moves in a straight line tangent to the circle at that point.

6. Uniform circular motion and speed

The word uniform means the speed is constant. It does not mean the velocity is constant. Velocity changes because direction changes.

This is a common mistake. If velocity were constant, there would be no acceleration. But in circular motion, there is always centripetal acceleration as long as the object is turning.

7. Steps for solving uniform circular motion problems

1. Identify the radius, speed, and mass.
2. Use $$a_c = \frac{v^2}{r}$$ to find centripetal acceleration if needed.
3. Use $$F_c = m\frac{v^2}{r}$$ to find the inward net force if needed.
4. Decide which real force is causing the inward force.
5. Check the direction: the centripetal net force must point toward the center.

Worked Example 1: Finding centripetal acceleration

A bicycle moves around a circular track of radius \(20\,\text{m}\) at a constant speed of \(10\,\text{m/s}\). Find the centripetal acceleration.

Step 1: Write the formula.

$$a_c = \frac{v^2}{r}$$

Step 2: Substitute the values.

$$a_c = \frac{(10)^2}{20}$$

$$a_c = \frac{100}{20} = 5\,\text{m/s}^2$$

Answer: The centripetal acceleration is \(5\,\text{m/s}^2\), directed toward the center of the track.

Worked Example 2: Finding centripetal force

A \(2.0\,\text{kg}\) ball is whirled in a horizontal circle of radius \(0.50\,\text{m}\) at a speed of \(4.0\,\text{m/s}\). What centripetal force is needed?

Step 1: Use the force formula.

$$F_c = m\frac{v^2}{r}$$

Step 2: Substitute the values.

$$F_c = 2.0\left(\frac{(4.0)^2}{0.50}\right)$$

$$F_c = 2.0\left(\frac{16}{0.50}\right)$$

$$F_c = 2.0(32) = 64\,\text{N}$$

Answer: The required centripetal force is \(64\,\text{N}\), toward the center. In this case, the tension in the string provides that force.

Worked Example 3: Friction providing centripetal force

A \(1000\,\text{kg}\) car turns on a flat circular road of radius \(50\,\text{m}\) at \(15\,\text{m/s}\). Find the centripetal force. What force provides it?

Step 1: Use the centripetal force formula.

$$F_c = m\frac{v^2}{r}$$

Step 2: Substitute the values.

$$F_c = 1000\left(\frac{15^2}{50}\right)$$

$$F_c = 1000\left(\frac{225}{50}\right)$$

$$F_c = 1000(4.5) = 4500\,\text{N}$$

Answer: The car needs an inward net force of \(4500\,\text{N}\). On a flat road, this force is provided by friction between the tires and the road.

Worked Example 4: Gravity as the centripetal force

A satellite of mass \(500\,\text{kg}\) moves in a circular orbit with radius \(7000\,\text{m}\) and speed \(140\,\text{m/s}\). Find the centripetal force.

Step 1: Use the formula.

$$F_c = m\frac{v^2}{r}$$

Step 2: Substitute.

$$F_c = 500\left(\frac{140^2}{7000}\right)$$

$$F_c = 500\left(\frac{19600}{7000}\right)$$

$$F_c = 500(2.8) = 1400\,\text{N}$$

Answer: The centripetal force is \(1400\,\text{N}\), and gravity provides this force.

8. Common mistakes to avoid

• Mistake 1: Thinking there is no acceleration because speed is constant.
  There is acceleration because direction changes.
• Mistake 2: Thinking centripetal force is a separate new force.
  It is actually the net inward force caused by real forces like tension, friction, or gravity.
• Mistake 3: Pointing the centripetal force in the direction of motion.
  The centripetal force points toward the center, not forward.
• Mistake 4: Forgetting to square the speed in the formula \(a_c = \frac{v^2}{r}\).

9. Key ideas to remember

• Uniform circular motion means motion in a circle at constant speed.
• An object in circular motion has centripetal acceleration toward the center.
• The formula for centripetal acceleration is $$a_c = \frac{v^2}{r}$$
• The formula for centripetal force is $$F_c = m\frac{v^2}{r}$$
• Centripetal force is provided by real forces such as tension, friction, or gravity.
• If the inward force disappears, the object moves in a straight tangent line.

Brief Summary

Uniform circular motion happens when an object moves in a circle at constant speed. Even though the speed stays the same, the object is accelerating because its direction changes. That acceleration is called centripetal acceleration, and it always points toward the center of the circle.

The formulas $$a_c = \frac{v^2}{r}$$ and $$F_c = m\frac{v^2}{r}$$ help calculate the acceleration and net inward force. The centripetal force is not a separate force; it is provided by real forces like tension, friction, or gravity depending on the situation.

Universal Gravitation is the idea that every object with mass pulls on every other object with mass. This pull is called gravity. Gravity is the force that keeps planets in orbit around the Sun, keeps the Moon in orbit around Earth, and pulls objects toward the ground.

Newton explained this with the law of universal gravitation. The word universal means it applies everywhere, not just on Earth. The same rule explains why an apple falls and why planets move through space.

The law says that the gravitational force between two objects depends on two main things:

• the masses of the objects
• the distance between their centers

The mathematical form of Newton's law is:

In this equation:

• \(F\) = gravitational force in newtons (N)
• \(G\) = universal gravitational constant
• \(m_1\) and \(m_2\) = the masses of the two objects in kilograms (kg)
• \(r\) = the distance between the centers of the objects in meters (m)

The value of the gravitational constant is:

This number is very small, which is why gravitational force between everyday objects is usually too weak to notice. But when one or both objects are very massive, like Earth or the Sun, gravity becomes very important.

How mass affects gravity: If one mass increases, the gravitational force increases. If both masses double, the force becomes 4 times larger because both masses are multiplied together.

How distance affects gravity: Gravity follows an inverse-square law. This means the force changes with the square of the distance.

• If the distance doubles, the force becomes \(\frac{1}{4}\) as large.
• If the distance triples, the force becomes \(\frac{1}{9}\) as large.
• If the distance is cut in half, the force becomes 4 times larger.

This inverse-square relationship is one of the most important ideas in gravitation. A small increase in distance can cause a large decrease in force.

Gravity near Earth is often described using the gravitational field strength, also called the acceleration due to gravity, \(g\). Near Earth's surface,

An object's weight is the gravitational force acting on it:

Weight depends on the local value of gravity, but mass does not change from place to place. For example, your mass is the same on Earth and on the Moon, but your weight would be smaller on the Moon because gravity there is weaker.

Newton's law can also explain why \(g\) depends on the mass and radius of a planet. For a planet of mass \(M\) and radius \(r\),

This means a planet with more mass usually has stronger gravity, but a larger radius can make gravity weaker at the surface because the surface is farther from the center.

Orbital motion happens when gravity continuously pulls an object inward while the object keeps moving forward. Instead of falling straight down, it keeps "falling around" the planet or star.

For a circular orbit, gravity provides the force needed to keep the object moving in a circle. This leads to the orbital speed formula:

In this equation:

• \(v\) = orbital speed
• \(G\) = gravitational constant
• \(M\) = mass of the planet or star being orbited
• \(r\) = distance from the center of that planet or star to the orbiting object

This formula shows that an object closer to a planet must move faster to stay in orbit. An object farther away can move more slowly.

We can also rearrange the orbital velocity formula to find the mass of a planet or star:

This is very useful in astronomy. By measuring how fast a moon or satellite moves and how far it is from the planet, scientists can calculate the planet's mass.

Important note about distance: In gravitation problems, the distance \(r\) is measured from center to center, not from the surfaces. For planets and moons, this matters a lot.

Steps for solving gravitation problems

1. Write down what is known and what must be found.
2. Choose the correct formula.
3. Make sure all units are in SI units: kilograms, meters, seconds.
4. Substitute carefully.
5. Check whether the answer makes sense. For example, a larger distance should usually give a smaller force.

Worked Example 1: Finding gravitational force between two objects

Suppose two objects have masses \(m_1 = 5.0\ \text{kg}\) and \(m_2 = 12.0\ \text{kg}\), and their centers are \(r = 2.0\ \text{m}\) apart. Find the gravitational force between them.

Use the formula:

Substitute the values:

Answer: The gravitational force is about \(1.0 \times 10^{-9}\ \text{N}\).

This is an extremely small force, which helps explain why gravity between small everyday objects is hard to notice.

Worked Example 2: How changing distance affects force

If the gravitational force between two objects is \(36\ \text{N}\) when they are distance \(r\) apart, what is the force when the distance becomes \(2r\)?

Because gravity follows an inverse-square law, doubling the distance makes the force one-fourth as large:

Answer: The new force is \(9\ \text{N}\).

This example shows that you do not always need to use the full formula. Sometimes recognizing the inverse-square pattern is faster.

Worked Example 3: Finding orbital speed of a satellite

A satellite orbits Earth at a distance of \(r = 7.0 \times 10^6\ \text{m}\) from Earth's center. Earth's mass is \(M = 5.97 \times 10^{24}\ \text{kg}\). Find the satellite's orbital speed.

Use:

Substitute:

First multiply in the numerator:

Now divide by \(r\):

Take the square root:

Answer: The orbital speed is about \(7.5 \times 10^3\ \text{m/s}\), or \(7500\ \text{m/s}\).

This is a very high speed, which is why satellites can remain in orbit instead of falling straight down.

Worked Example 4: Finding the mass of a planet from an orbiting moon

A moon travels in a circular orbit around a planet with speed \(v = 4.0 \times 10^3\ \text{m/s}\). Its orbital radius is \(r = 2.0 \times 10^7\ \text{m}\). Find the mass of the planet.

Use:

Substitute the values:

Square the speed:

Multiply by \(r\):

Now divide by \(G\):

Answer: The planet's mass is about \(4.8 \times 10^{24}\ \text{kg}\).

This method is one of the main ways scientists determine the masses of planets and stars.

Common mistakes to avoid

• Using the distance between surfaces instead of the distance between centers.
• Forgetting to square the distance in the denominator.
• Confusing mass and weight.
• Using kilometers instead of meters without converting.
• Leaving out units in the final answer.

Key ideas to remember

• Every mass attracts every other mass.
• Gravitational force increases with mass.
• Gravitational force decreases with the square of the distance.
• Weight is the gravitational force on an object.
• Gravity is the force that keeps satellites, moons, and planets in orbit.
• Orbital speed depends on the mass of the object being orbited and the orbital radius.

Brief Summary

Newton's law of universal gravitation explains that all objects with mass attract one another with a force given by $$F = G\frac{m_1m_2}{r^2}$$. This force gets stronger with larger masses and much weaker as distance increases because of the inverse-square law. The same idea explains weight, gravity at a planet's surface, and the motion of satellites and planets in orbit. By using formulas such as $$v = \sqrt{\frac{GM}{r}}$$ and $$M = \frac{v^2r}{G}$$, we can calculate orbital speeds and even find the mass of planets.

Work and Power are two important ideas in mechanics. They help us describe how forces transfer energy and how quickly that energy is transferred.

In everyday language, the word “work” can mean any effort. In science, work has a very specific meaning. Work happens when a force causes an object to move in the direction of the force.

Power tells us how fast work is done or how fast energy is transferred. Two people might do the same amount of work, but the one who does it in less time uses more power.

Understanding work and power helps explain many real-life situations, such as lifting boxes, pushing carts, climbing stairs, and how engines and machines perform.

1. What is Work?

In science, work is done when:

• a force acts on an object, and
• the object moves in the direction of that force.

If the force and motion are parallel, the formula for work is:

where:

• W = work done, measured in joules (J)
• F = force, measured in newtons (N)
• d = displacement, measured in meters (m)

One joule is equal to one newton-meter:

This formula works when the force is in the same direction as the displacement.

2. When is Work Done?

It is important to know that not every force does work.

Work is done when a force causes movement in its direction. For example, if you push a box across the floor and it moves forward, you do work on the box.

No work is done if there is no displacement. If you push on a wall and it does not move, then:

Even though you may feel tired, no scientific work is done on the wall because the wall does not move.

No work is also done if the force is not in the direction of motion. In this lesson, we focus only on forces parallel to displacement.

3. Positive, Negative, and Zero Work

Work can be positive, negative, or zero depending on the direction of the force compared with the displacement.

• Positive work: the force and displacement are in the same direction.
• Negative work: the force is opposite the displacement.
• Zero work: there is no displacement, or the force does not act along the displacement in this simplified lesson.

For example, if you lift a book upward, the upward force you apply does positive work. If friction acts backward while the book slides forward, friction does negative work.

4. Work and Energy

Work is closely connected to energy. When work is done on an object, energy is transferred.

For example:

• Lifting an object gives it more gravitational potential energy.
• Pushing a moving object can increase its kinetic energy.
• Friction can remove mechanical energy and transfer it as heat.

This means work is one way that energy moves from one object or system to another.

5. What is Power?

Power is the rate at which work is done or energy is transferred.

The formula for power is:

where:

• P = power, measured in watts (W)
• W = work, measured in joules (J)
• t = time, measured in seconds (s)

One watt means one joule of work is done each second:

A larger power means work is being done more quickly.

6. Work vs. Power

Students often confuse work and power, so it helps to compare them clearly.

• Work tells how much energy is transferred.
• Power tells how fast that energy is transferred.

For example, two students may both lift the same box to the same height, so they do the same work. But if one student lifts it in 2 seconds and the other takes 5 seconds, the first student uses more power.

7. Steps for Solving Work and Power Problems

When solving problems, follow these steps:

1. Identify what the question is asking.
2. Write down the known values and units.
3. Choose the correct formula: either \(W = Fd\) or \(P = \frac{W}{t}\).
4. Substitute the values carefully.
5. Calculate and include the correct unit.
6. Check whether the answer makes sense.

8. Worked Examples

Example 1: Basic Work

A student pushes a box with a force of \(20 ext{ N}\) across the floor for \(5 ext{ m}\). The force and displacement are in the same direction. How much work is done?

Step 1: Write the formula

Step 2: Substitute the values

Step 3: Calculate

Answer: The work done is 100 J.

Example 2: No Displacement

A person pushes on a heavy crate with a force of \(50 ext{ N}\), but the crate does not move. How much work is done on the crate?

Step 1: Use the formula

Step 2: Since the displacement is zero, \(d = 0\)

Step 3: Calculate

Answer: The work done is 0 J.

This is a key idea: if there is no movement, there is no work in the scientific sense.

Example 3: Finding Power

A machine does \(600 ext{ J}\) of work in \(12 ext{ s}\). What is its power?

Step 1: Write the formula

Step 2: Substitute the values

Step 3: Calculate

Answer: The power of the machine is 50 W.

Example 4: Work and Then Power

A worker lifts a box upward with a force of \(150 ext{ N}\) through a height of \(2 ext{ m}\). It takes \(3 ext{ s}\). Find:

1. the work done
2. the power used

Part A: Find the work

Part B: Find the power

Answer:

• Work done = 300 J
• Power used = 100 W

9. Common Mistakes to Avoid

• Forgetting displacement: A force alone does not mean work is done. The object must move.
• Mixing up work and power: Work is energy transferred; power is the rate of transfer.
• Using the wrong unit: Work is measured in joules, while power is measured in watts.
• Ignoring time in power questions: If a question asks for power, time must be included.
• Not checking direction: In this lesson, use \(W = Fd\) only when force and displacement are parallel.

10. Real-Life Applications

Work and power appear in many everyday situations:

• Climbing stairs: You do work against gravity. If you run up the stairs, your power is greater.
• Car engines: Engines do work to move a car. More powerful engines can do work faster.
• Elevators: Elevators do work lifting people and objects upward.
• Sports: A weightlifter does work lifting a barbell. Lifting it faster means greater power.

11. Quick Check for Understanding

Ask yourself these questions:

• If I push an object and it does not move, is work done?
• If the same amount of work is done in less time, is power greater or smaller?
• What are the units for work and power?
• When can I use \(W = Fd\)?

The answers should be:

• No, because there is no displacement.
• Power is greater.
• Work is measured in joules; power is measured in watts.
• When force and displacement are parallel.

12. Summary

Work is the transfer of energy when a force moves an object in the direction of the force. For parallel force and displacement, use:

Power tells how quickly work is done:

Remember:

• Work needs both force and displacement.
• No displacement means no work.
• Power depends on time.
• Work is measured in joules, and power is measured in watts.

These ideas help us understand how energy is transferred and how efficiently people and machines do tasks.

Lesson: Kinetic and Potential Energy

Energy is one of the most important ideas in science. It helps us describe how objects move, how they can cause change, and how motion and position are connected.

In this lesson, you will learn about two major forms of mechanical energy: kinetic energy and potential energy. You will also learn how to calculate them and how they change when an object moves.

Kinetic energy is the energy an object has because it is moving. A faster object has more kinetic energy than a slower object of the same mass. A heavier object also has more kinetic energy than a lighter object moving at the same speed.

The formula for translational kinetic energy is

$$KE = \frac{1}{2}mv^2$$

In this formula:

• KE is kinetic energy, measured in joules (J)
• m is mass, measured in kilograms (kg)
• v is speed, measured in meters per second (m/s)

A joule is the standard unit of energy. One joule is a small amount of energy, but it is the unit used for all of the energy calculations in this lesson.

Notice that speed is squared in the kinetic energy formula. This means speed has a very strong effect on kinetic energy. If speed doubles, kinetic energy becomes four times as large.

Potential energy is stored energy. It depends on an objects position, shape, or condition. In 10th Grade science, two common types are gravitational potential energy and elastic potential energy.

Gravitational potential energy is the energy stored because of an objects height above the ground or another chosen reference level. The higher the object is, the more gravitational potential energy it has.

The formula for gravitational potential energy is

$$GPE = mgh$$

In this formula:

• GPE is gravitational potential energy in joules (J)
• m is mass in kilograms (kg)
• g is the gravitational field strength, usually \(9.8\,m/s^2\) on Earth
• h is height in meters (m)

This formula shows that gravitational potential energy increases when mass increases or height increases.

Elastic potential energy is stored when an object is stretched or compressed. A spring, rubber band, or bow can store elastic potential energy.

The formula for elastic potential energy is

$$EPE = \frac{1}{2}kx^2$$

In this formula:

• EPE is elastic potential energy in joules (J)
• k is the spring constant, which tells how stiff the spring is
• x is the distance stretched or compressed from the springs normal length, measured in meters

Like kinetic energy, the stretch distance is squared. This means that a larger stretch causes a much larger increase in elastic potential energy.

Mechanical energy is the total of kinetic energy and potential energy. For many situations, we write

$$ME = KE + PE$$

If there is little or no energy lost to friction, air resistance, or heat, then the total mechanical energy stays the same. This is called conservation of energy.

That means energy can change form, but it is not created or destroyed. For example, when a ball falls, its gravitational potential energy decreases while its kinetic energy increases.

Here is the basic idea of energy changes in motion:

• High position, low speed: more potential energy, less kinetic energy
• Low position, high speed: less potential energy, more kinetic energy
• Stopped spring: no elastic potential energy if it is not stretched or compressed
• Stretched spring: more elastic potential energy

Choosing a reference level for gravitational potential energy is important. Height is measured from some chosen zero point. In many problems, the ground is used as \(h=0\). The exact zero point can change, but the energy changes are what matter most.

Worked Example 1: Finding Kinetic Energy

A \(4.0\,kg\) cart moves at \(3.0\,m/s\). Find its kinetic energy.

Step 1: Write the formula.

$$KE = \frac{1}{2}mv^2$$

Step 2: Substitute the values.

$$KE = \frac{1}{2}(4.0)(3.0)^2$$

Step 3: Calculate.

$$KE = 0.5 \times 4.0 \times 9.0 = 18\,J$$

Answer: The cart has 18 J of kinetic energy.

This example shows how strongly speed affects kinetic energy. If the same cart moved at \(6.0\,m/s\), its kinetic energy would not just double. It would become four times greater because speed is squared.

Worked Example 2: Finding Gravitational Potential Energy

A \(2.0\,kg\) book is held \(1.5\,m\) above the floor. Find its gravitational potential energy relative to the floor.

Step 1: Write the formula.

$$GPE = mgh$$

Step 2: Substitute the values.

$$GPE = (2.0)(9.8)(1.5)$$

Step 3: Calculate.

$$GPE = 29.4\,J$$

Answer: The book has 29.4 J of gravitational potential energy.

If the book is lifted higher, its gravitational potential energy increases. If it is lowered to the floor, its gravitational potential energy relative to the floor becomes zero.

Worked Example 3: Energy Change During a Fall

A \(1.0\,kg\) ball is dropped from a height of \(5.0\,m\). Ignore air resistance. What is its speed just before it hits the ground?

Step 1: Find the starting gravitational potential energy.

$$GPE = mgh = (1.0)(9.8)(5.0) = 49\,J$$

At the top, the ball is at rest, so its kinetic energy is \(0\,J\). Its total mechanical energy is \(49\,J\).

Step 2: Use conservation of energy.

Just before the ball hits the ground, its height is about zero, so its gravitational potential energy is about \(0\,J\). That means all \(49\,J\) has changed into kinetic energy.

$$KE = 49\,J$$

Step 3: Solve for speed using the kinetic energy formula.

$$\frac{1}{2}mv^2 = 49$$

$$\frac{1}{2}(1.0)v^2 = 49$$

$$0.5v^2 = 49$$

$$v^2 = 98$$

$$v \approx 9.9\,m/s$$

Answer: The balls speed just before hitting the ground is about \(9.9\,m/s\).

This example shows how gravitational potential energy can turn into kinetic energy as an object falls.

Worked Example 4: Elastic Potential Energy in a Spring

A spring has a spring constant of \(200\,N/m\). It is compressed by \(0.10\,m\). How much elastic potential energy is stored?

Step 1: Write the formula.

$$EPE = \frac{1}{2}kx^2$$

Step 2: Substitute the values.

$$EPE = \frac{1}{2}(200)(0.10)^2$$

Step 3: Calculate.

$$EPE = 0.5 \times 200 \times 0.01 = 1.0\,J$$

Answer: The spring stores 1.0 J of elastic potential energy.

If the spring were compressed twice as far, the stored energy would become four times greater because the distance is squared.

Comparing the Three Energy Equations

• Kinetic energy: $$KE = \frac{1}{2}mv^2$$
• Gravitational potential energy: $$GPE = mgh$$
• Elastic potential energy: $$EPE = \frac{1}{2}kx^2$$

Each equation depends on different quantities:

• Kinetic energy depends on mass and speed.
• Gravitational potential energy depends on mass, gravity, and height.
• Elastic potential energy depends on spring stiffness and amount of stretch or compression.

Common mistakes to avoid

• Do not forget to square \(v\) in kinetic energy.
• Do not forget to square \(x\) in elastic potential energy.
• Use meters, kilograms, and meters per second in the formulas.
• Make sure height is measured from the chosen reference level.
• Remember that energy is measured in joules.

How these ideas connect to real life

• A roller coaster has high gravitational potential energy at the top of a hill and high kinetic energy at the bottom.
• A moving car has kinetic energy because of its speed.
• A drawn bow stores elastic potential energy that becomes kinetic energy when the arrow is released.
• A falling apple changes gravitational potential energy into kinetic energy.

Quick Review

1. Kinetic energy is the energy of motion: $$KE = \frac{1}{2}mv^2$$
2. Gravitational potential energy is stored because of height: $$GPE = mgh$$
3. Elastic potential energy is stored in stretched or compressed springs: $$EPE = \frac{1}{2}kx^2$$
4. Total mechanical energy is the sum of kinetic and potential energy.
5. When no significant energy is lost, mechanical energy is conserved.

Brief Summary

Kinetic energy depends on how much mass an object has and how fast it moves. Potential energy is stored energy, and two important types are gravitational potential energy and elastic potential energy.

As objects move, energy can change from one form to another. A falling object changes gravitational potential energy into kinetic energy, and a spring can store elastic potential energy that later becomes motion. By using the correct formulas and units, you can calculate and compare these energies in many real situations.

Conservation of Mechanical Energy is one of the most useful ideas in mechanics. It helps us understand how moving objects speed up, slow down, rise, and fall by tracking how energy changes form.

In this lesson, you will learn what kinetic energy and potential energy are, when mechanical energy is conserved, how the Work-Energy Theorem connects force and motion, and what changes when friction or other non-conservative forces are present.

By the end, you should be able to decide whether mechanical energy stays constant, write the correct energy equation, and solve for unknown speeds, heights, or energy changes.

1. What is mechanical energy?

Mechanical energy is the total energy an object has because of its motion and position. It is the sum of:

• Kinetic energy: energy of motion
• Potential energy: stored energy due to position or arrangement

So we write:

$$E_{mech} = KE + PE$$

For most 10th Grade problems, the two main forms are:

• Gravitational potential energy: $$PE_g = mgh$$
• Kinetic energy: $$KE = \frac{1}{2}mv^2$$

In these equations:

• \(m\) = mass in kilograms
• \(g\) = gravitational field strength, about \(9.8\,m/s^2\) on Earth
• \(h\) = height in meters
• \(v\) = speed in meters per second

If an object is high above the ground, it has more gravitational potential energy. If it is moving fast, it has more kinetic energy.

2. The idea of conservation

To conserve something means the total amount stays the same. In a system where only conservative forces act, the total mechanical energy remains constant.

That means:

$$KE_i + PE_i = KE_f + PE_f$$

The subscripts \(i\) and \(f\) mean initial and final.

This does not mean kinetic energy stays the same or potential energy stays the same. Instead, one form can change into the other while the total stays constant.

For example:

• As a ball falls, gravitational potential energy decreases.
• At the same time, kinetic energy increases.
• If we ignore air resistance, the total mechanical energy stays the same.

3. Conservative and non-conservative forces

A conservative force transfers energy between kinetic and potential forms without removing energy from the mechanical system. Gravity is the main example you will use here.

A non-conservative force changes mechanical energy into other forms, such as thermal energy or sound. The most common example is friction.

This leads to two important cases:

• Frictionless system: mechanical energy is conserved.
• System with friction or air resistance: mechanical energy is not conserved, even though total energy is still conserved overall.

So in a real system with friction, some mechanical energy is transformed into thermal energy.

4. The Work-Energy Theorem

The Work-Energy Theorem says that the net work done on an object equals its change in kinetic energy:

$$W_{net} = \Delta KE = KE_f - KE_i$$

This theorem is very important because it connects forces and energy.

If the net work is positive, the object gains kinetic energy and usually speeds up. If the net work is negative, the object loses kinetic energy and usually slows down.

When only gravity does work, the work done by gravity changes potential energy into kinetic energy or kinetic energy into potential energy. In that case, mechanical energy is conserved.

When friction also does work, we include its effect as non-conservative work:

$$W_{nc} = \Delta E_{mech} = (KE_f + PE_f) - (KE_i + PE_i)$$

Here, \(W_{nc}\) means work done by non-conservative forces.

If friction removes energy, \(W_{nc}\) is negative.

5. Energy transformations in common situations

Object falling downward

• \(PE\) decreases
• \(KE\) increases
• Total mechanical energy stays constant if there is no air resistance

Object moving upward

• \(KE\) decreases
• \(PE\) increases
• Total mechanical energy stays constant if there is no friction or air resistance

Object sliding on a rough surface

• Mechanical energy decreases
• Some energy becomes thermal energy because of friction

6. Choosing a reference height

Potential energy depends on where you choose \(h=0\). This is called the reference level.

You may choose the ground, the floor, or the lowest point in the problem as zero height. What matters is being consistent throughout the problem.

Even if your zero point changes, the final answer will still be correct if all heights are measured from the same reference.

7. How to solve conservation of mechanical energy problems

1. Identify the initial and final positions.
2. Decide whether the system is frictionless or has non-conservative forces.
3. Write expressions for \(KE\) and \(PE\) at each point.
4. Use either:
   • $$KE_i + PE_i = KE_f + PE_f$$ for frictionless situations, or
   • $$W_{nc} = (KE_f + PE_f) - (KE_i + PE_i)$$ when friction is present.
5. Substitute values and solve carefully.
6. Check if the answer makes sense physically.

8. Worked Example 1: Falling object in a frictionless system

A \(2.0\,kg\) ball is dropped from rest from a height of \(5.0\,m\). What is its speed just before it hits the ground? Ignore air resistance.

Step 1: Write known information.

• \(m = 2.0\,kg\)
• \(h_i = 5.0\,m\)
• \(v_i = 0\,m/s\)
• \(h_f = 0\,m\)

Step 2: Use conservation of mechanical energy.

$$KE_i + PE_i = KE_f + PE_f$$

Since the ball starts from rest, \(KE_i = 0\). At the ground, \(PE_f = 0\).

So:

$$mgh_i = \frac{1}{2}mv_f^2$$

Step 3: Substitute values.

$$2.0(9.8)(5.0) = \frac{1}{2}(2.0)v_f^2$$

$$98 = v_f^2$$

$$v_f = \sqrt{98} \approx 9.9\,m/s$$

Answer: The ball hits the ground at about \(9.9\,m/s\).

Important idea: The mass canceled out. In a frictionless fall, speed depends on height, not mass.

9. Worked Example 2: Roller coaster going downhill

A roller coaster car starts from rest at the top of a hill \(20\,m\) high. What is its speed at the bottom if friction is ignored?

Step 1: Identify energies.

• At the top: mostly gravitational potential energy
• At the bottom: mostly kinetic energy

Use:

$$KE_i + PE_i = KE_f + PE_f$$

Since it starts from rest, \(KE_i = 0\). At the bottom, take \(PE_f = 0\).

$$mgh = \frac{1}{2}mv^2$$

Cancel \(m\):

$$gh = \frac{1}{2}v^2$$

$$v^2 = 2gh$$

$$v = \sqrt{2(9.8)(20)}$$

$$v = \sqrt{392} \approx 19.8\,m/s$$

Answer: The speed at the bottom is about \(19.8\,m/s\).

10. Worked Example 3: Thrown upward

A \(0.50\,kg\) ball is thrown straight upward with an initial speed of \(12\,m/s\). How high does it rise before stopping for an instant? Ignore air resistance.

Step 1: Think about the motion.

At the highest point, the ball’s speed is zero for an instant. So its kinetic energy becomes zero there. Its initial kinetic energy changes into gravitational potential energy.

Step 2: Use conservation of mechanical energy.

$$KE_i + PE_i = KE_f + PE_f$$

Choose the launch point as \(h=0\), so \(PE_i = 0\). At the top, \(KE_f = 0\).

$$\frac{1}{2}mv_i^2 = mgh_f$$

Substitute values:

$$\frac{1}{2}(0.50)(12^2) = (0.50)(9.8)h_f$$

$$36 = 4.9h_f$$

$$h_f = \frac{36}{4.9} \approx 7.35\,m$$

Answer: The ball rises about \(7.35\,m\).

11. Worked Example 4: Friction removes mechanical energy

A \(3.0\,kg\) box slides across a horizontal floor. Its initial speed is \(8.0\,m/s\), and it comes to rest because of friction. How much mechanical energy is lost?

Step 1: Identify the energy change.

Because the floor is horizontal, the height does not change. So potential energy stays the same. The box loses its kinetic energy, and that energy is transformed mainly into thermal energy due to friction.

Step 2: Calculate initial kinetic energy.

$$KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(3.0)(8.0)^2$$

$$KE_i = 1.5 \times 64 = 96\,J$$

Step 3: Final kinetic energy.

Since the box stops, \(KE_f = 0\).

Step 4: Find change in mechanical energy.

$$\Delta E_{mech} = E_f - E_i = 0 - 96 = -96\,J$$

The negative sign means the system lost mechanical energy.

Answer: The box loses \(96\,J\) of mechanical energy.

This also means friction did \(-96\,J\) of work on the box.

12. Comparing frictionless and dissipative systems

A frictionless system is an ideal situation where no energy is lost to friction or air resistance. In these problems:

• Mechanical energy stays constant.
• Use $$KE_i + PE_i = KE_f + PE_f$$

A dissipative system includes forces like friction that reduce mechanical energy. In these problems:

• Mechanical energy decreases.
• Total energy is still conserved, but some becomes thermal or sound energy.
• Use $$W_{nc} = (KE_f + PE_f) - (KE_i + PE_i)$$

13. Common mistakes to avoid

• Mixing up total mechanical energy with one type of energy. Kinetic and potential energy can change, even when total mechanical energy stays constant.
• Forgetting to choose a reference height. You need a clear zero level for gravitational potential energy.
• Using conservation of mechanical energy when friction is present. If friction matters, mechanical energy is not conserved by itself.
• Forgetting units. Energy is measured in joules \((J)\).
• Ignoring that speed is squared in kinetic energy. Doubling speed makes kinetic energy four times larger.

14. Quick check for understanding

Ask yourself these questions when solving a problem:

• Is the object changing height, speed, or both?
• Is friction or air resistance present?
• Which energies exist at the start and end?
• Should I use conservation of mechanical energy or include non-conservative work?

15. Lesson summary

Mechanical energy is the sum of kinetic and potential energy:

$$E_{mech} = KE + PE$$

In a frictionless system, mechanical energy is conserved:

$$KE_i + PE_i = KE_f + PE_f$$

The Work-Energy Theorem states:

$$W_{net} = \Delta KE$$

When friction or other non-conservative forces are present, mechanical energy changes because some energy is transformed into thermal energy or sound:

$$W_{nc} = (KE_f + PE_f) - (KE_i + PE_i)$$

Understanding these ideas helps you describe motion without always using force equations. Instead, you can follow how energy moves from one form to another.

Momentum and Impulse

When objects move, they carry more than just speed. A fast-moving truck is much harder to stop than a fast-moving bicycle. This is because moving objects have momentum.

Momentum helps us describe motion in a way that includes both mass and velocity. Impulse helps us understand how a force changes an object’s momentum, especially during collisions and safety situations like airbags, helmets, and seatbelts.

In this lesson, you will learn what momentum and impulse are, how to calculate them, and how they are used to explain force mitigation in real life.

1. What is momentum?

Momentum is the quantity of motion an object has. It depends on two things:

• the object’s mass
• the object’s velocity

The formula for momentum is:

where:

• \(p\) = momentum
• \(m\) = mass in kilograms (kg)
• \(v\) = velocity in meters per second (m/s)

The unit for momentum is:

Because velocity has direction, momentum also has direction. That means momentum is a vector. If an object moves to the right, its momentum is to the right. If it moves to the left, its momentum is to the left.

2. How mass and velocity affect momentum

An object with more mass has more momentum if it is moving at the same velocity. An object moving faster also has more momentum if its mass stays the same.

For example:

• A bowling ball rolling slowly can still have large momentum because it has a lot of mass.
• A baseball moving very fast can also have noticeable momentum because its velocity is high.

This is why both mass and velocity matter when thinking about how hard something is to stop.

3. What is impulse?

Impulse describes how a force changes an object’s momentum. It depends on:

• the force applied
• the amount of time the force acts

The formula for impulse is:

where:

• \(J\) = impulse
• \(F\) = force in newtons (N)
• \(\Delta t\) = time interval in seconds (s)

The unit for impulse is:

This unit is equal to kg·m/s, which matches the unit of momentum.

4. The impulse-momentum theorem

The most important idea in this lesson is that impulse equals the change in momentum.

This can also be written as:

where:

• \(v_i\) = initial velocity
• \(v_f\) = final velocity

This equation tells us that if an object’s momentum changes, a force must have acted for some amount of time.

5. Why stopping time matters

In many real-world situations, the change in momentum is fixed. For example, if a car moving forward comes to a stop, its momentum changes from some positive value to zero.

If the momentum change is the same, then the impulse is the same. But the force can still be different depending on how long the stopping takes.

From the equation

if \(\Delta p\) stays the same, then increasing \(\Delta t\) makes the force smaller.

This is the key idea behind force mitigation in safety engineering.

6. Safety engineering: airbags, seatbelts, and helmets

Safety devices do not remove momentum instantly. Instead, they increase the time over which a person stops moving.

• Airbags increase the time it takes for a passenger to come to rest.
• Seatbelts help spread the stopping force and prevent sudden impact with the dashboard or windshield.
• Helmets add padding, which increases impact time and lowers the force on the head.

Because the stopping time is longer, the force is reduced. A smaller force means less chance of injury.

7. Positive and negative momentum

Since momentum depends on velocity, direction matters. You can choose one direction as positive and the opposite direction as negative.

For example:

• right might be positive
• left might be negative

If an object moving right slows down, its momentum change is negative. If an object moving left stops, its momentum change may be positive, depending on the sign convention.

Always pay attention to direction when solving momentum problems.

8. Steps for solving momentum and impulse problems

1. Identify the known values: mass, initial velocity, final velocity, force, or time.
2. Choose the correct formula: \(p = mv\), \(J = F\Delta t\), or \(F\Delta t = \Delta p\).
3. Make sure units are correct: kg, m/s, s, and N.
4. Include direction by using positive and negative signs if needed.
5. Substitute the numbers carefully.
6. Check whether the answer makes sense physically.

Worked Example 1: Finding momentum

A 4 kg cart moves to the right at 3 m/s. What is its momentum?

Step 1: Use the formula

Step 2: Substitute values

Answer: The cart’s momentum is 12 kg·m/s to the right.

Worked Example 2: Finding change in momentum and impulse

A 0.50 kg soccer ball is initially at rest. A player kicks it, and it moves forward at 8 m/s. What are the ball’s change in momentum and the impulse delivered?

Step 1: Write the momentum change formula

Step 2: Substitute values

Since impulse equals change in momentum:

Answer: The change in momentum is 4 kg·m/s forward, and the impulse is 4 N·s forward.

Worked Example 3: Finding force from impulse

A 0.20 kg tennis ball hits a racket and experiences an impulse of 3.0 N·s. If the contact time is 0.015 s, what average force acts on the ball?

Step 1: Use the impulse formula

Step 2: Solve for force

Step 3: Substitute values

Answer: The average force on the ball is 200 N.

Worked Example 4: Airbag and force mitigation

A passenger with mass 60 kg is moving at 12 m/s with a car. In a crash, the passenger comes to rest. Compare the average force if the stopping time is 0.10 s without an airbag and 0.50 s with an airbag.

Step 1: Find the change in momentum

Initial momentum:

Final momentum:

So the change in momentum is:

The negative sign means the momentum decreased. For force size, we compare magnitudes.

Step 2: Without airbag

Using magnitude:

Step 3: With airbag

Answer:

• Without airbag: 7200 N
• With airbag: 1440 N

The airbag increases stopping time, so the average force is much smaller.

9. Common mistakes to avoid

• Confusing speed and velocity: Momentum uses velocity, so direction matters.
• Forgetting units: Use kg for mass and m/s for velocity.
• Ignoring signs: Left and right may need negative and positive values.
• Mixing up momentum and force: Momentum is not the same as force. Force changes momentum.
• Forgetting the role of time: A longer impact time means a smaller force for the same momentum change.

10. Key ideas to remember

• Momentum is the product of mass and velocity: \(p = mv\).
• Impulse is force times time: \(J = F\Delta t\).
• Impulse equals change in momentum: \(J = \Delta p\).
• For the same change in momentum, increasing stopping time reduces force.
• This is why airbags, helmets, and padding help protect people.

Brief Summary

Momentum describes how hard it is to stop a moving object and depends on mass and velocity. Impulse describes how a force acting over time changes momentum. The impulse-momentum theorem, \(F\Delta t = \Delta p\), explains why increasing the time of impact lowers the force, which is the basic idea behind many safety devices.

Conservation of Momentum in Collisions is one of the most useful ideas in mechanics. It helps us predict what happens when objects crash, bounce, or stick together.

Momentum is the quantity of motion an object has. It depends on both mass and velocity. The formula is:

$$p = mv$$

Here, \(p\) is momentum, \(m\) is mass, and \(v\) is velocity.

Because velocity includes direction, momentum also has direction. In one-dimensional problems, we often use positive and negative signs to show direction. For example, motion to the right might be positive, while motion to the left is negative.

The big idea of this lesson is the law of conservation of momentum. It states that in an isolated system, the total momentum before a collision equals the total momentum after the collision.

$$\text{total momentum before} = \text{total momentum after}$$

Or, more mathematically:

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

The subscripts \(i\) and \(f\) mean initial and final.

This law works when the system is isolated, meaning no important outside forces change the total momentum during the collision. In real life, this is often approximately true if the collision happens quickly and friction or other outside forces are small.

Why momentum is conserved can be understood by thinking about forces during a collision. When object A pushes on object B, object B pushes back on object A with an equal and opposite force. These internal forces change the objects' individual momenta, but the total momentum of the system stays the same.

Important idea: Momentum is always conserved in collisions if the system is isolated. However, kinetic energy is not always conserved.

That is why collisions are divided into types.

1. Elastic collisions

• Momentum is conserved.
• Kinetic energy is also conserved.
• Objects bounce off each other without lasting deformation or heat loss in the model.

2. Inelastic collisions

• Momentum is conserved.
• Kinetic energy is not conserved.
• Some kinetic energy changes into sound, heat, or deformation.

3. Perfectly inelastic collisions

• Momentum is conserved.
• The objects stick together after the collision.
• This is a special type of inelastic collision.

In all three cases, the key rule is the same: total momentum before = total momentum after.

Working with 1D collisions

In one-dimensional motion, all velocities are along a single straight line. This makes momentum problems easier because we only need one equation for conservation of momentum.

To solve a 1D collision problem:

1. Choose a positive direction.
2. Write the momentum of each object, including signs.
3. Add the momenta before the collision.
4. Set that equal to the total momentum after the collision.
5. Solve for the unknown velocity or mass.

Sign convention matters. If an object moves in the negative direction, its velocity is negative, so its momentum is also negative.

Worked Example 1: Simple 1D collision

A \(2.0\text{ kg}\) cart moves to the right at \(4.0\text{ m/s}\). It collides with a \(3.0\text{ kg}\) cart at rest. After the collision, the \(2.0\text{ kg}\) cart moves at \(1.0\text{ m/s}\) to the right. Find the final velocity of the \(3.0\text{ kg}\) cart.

Step 1: Write what is known.

• \(m_1 = 2.0\text{ kg}\)
• \(v_{1i} = +4.0\text{ m/s}\)
• \(m_2 = 3.0\text{ kg}\)
• \(v_{2i} = 0\text{ m/s}\)
• \(v_{1f} = +1.0\text{ m/s}\)
• \(v_{2f} = ?\)

Step 2: Use conservation of momentum.

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

Substitute the values:

$$2.0(4.0) + 3.0(0) = 2.0(1.0) + 3.0v_{2f}$$

$$8.0 = 2.0 + 3.0v_{2f}$$

$$6.0 = 3.0v_{2f}$$

$$v_{2f} = 2.0\text{ m/s}$$

Answer: The \(3.0\text{ kg}\) cart moves to the right at \(2.0\text{ m/s}\).

Worked Example 2: Perfectly inelastic collision

A \(1.5\text{ kg}\) ball rolls to the right at \(6.0\text{ m/s}\). It collides with a \(0.5\text{ kg}\) ball rolling to the left at \(2.0\text{ m/s}\). The balls stick together. Find their final velocity.

Step 1: Choose right as positive.

• \(m_1 = 1.5\text{ kg}\)
• \(v_{1i} = +6.0\text{ m/s}\)
• \(m_2 = 0.5\text{ kg}\)
• \(v_{2i} = -2.0\text{ m/s}\)

Because they stick together, they have one shared final velocity, \(v_f\).

Step 2: Apply conservation of momentum.

$$m_1v_{1i} + m_2v_{2i} = (m_1+m_2)v_f$$

$$1.5(6.0) + 0.5(-2.0) = (1.5+0.5)v_f$$

$$9.0 - 1.0 = 2.0v_f$$

$$8.0 = 2.0v_f$$

$$v_f = 4.0\text{ m/s}$$

Answer: The stuck-together balls move to the right at \(4.0\text{ m/s}\).

This is a perfectly inelastic collision because the objects stick together. Momentum is conserved, but some kinetic energy is lost to other forms such as sound, heat, or deformation.

Elastic collisions in 1D

In an elastic collision, both momentum and kinetic energy are conserved. For many 10th Grade problems, you may be asked mainly to use momentum conservation and identify that kinetic energy also stays the same.

Kinetic energy is:

$$KE = \frac{1}{2}mv^2$$

In elastic collisions, the total kinetic energy before equals the total kinetic energy after:

$$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$$

You do not always need this equation unless the problem specifically says the collision is elastic and asks for more than one unknown.

Worked Example 3: Recognizing collision type

A \(2.0\text{ kg}\) object moves at \(5.0\text{ m/s}\) and hits a \(2.0\text{ kg}\) object at rest. After the collision, the first object stops and the second object moves at \(5.0\text{ m/s}\).

Step 1: Check momentum before and after.

Before:

$$p_i = 2.0(5.0) + 2.0(0) = 10\text{ kg·m/s}$$

After:

$$p_f = 2.0(0) + 2.0(5.0) = 10\text{ kg·m/s}$$

Momentum is conserved.

Step 2: Check kinetic energy.

Before:

$$KE_i = \frac{1}{2}(2.0)(5.0)^2 = 25\text{ J}$$

After:

$$KE_f = \frac{1}{2}(2.0)(5.0)^2 = 25\text{ J}$$

Kinetic energy is also conserved.

Answer: This is an elastic collision.

Moving from 1D to 2D collisions

In two-dimensional collisions, objects can move in different directions after impact. Momentum is still conserved, but now we must think about horizontal and vertical directions separately.

The total momentum in the \(x\)-direction is conserved, and the total momentum in the \(y\)-direction is also conserved.

That means we use two equations:

$$\sum p_x\text{ before} = \sum p_x\text{ after}$$

$$\sum p_y\text{ before} = \sum p_y\text{ after}$$

This is because momentum is a vector, and vectors have components.

How to solve 2D momentum problems

1. Draw the situation and label directions.
2. Choose \(x\)- and \(y\)-axes.
3. Break each momentum into \(x\)- and \(y\)-components if needed.
4. Apply momentum conservation separately in each direction.
5. Solve the equations.

If an object is moving only horizontally, then its \(y\)-momentum is zero. If it is moving only vertically, then its \(x\)-momentum is zero.

Worked Example 4: Simple 2D collision

A \(2.0\text{ kg}\) puck moves east at \(3.0\text{ m/s}\). A \(1.0\text{ kg}\) puck moves north at \(4.0\text{ m/s}\). They collide and stick together. Find the velocity of the combined puck after the collision.

Step 1: Find momentum in each direction before the collision.

Take east as positive \(x\), north as positive \(y\).

In the \(x\)-direction:

$$p_x = 2.0(3.0) + 1.0(0) = 6.0\text{ kg·m/s}$$

In the \(y\)-direction:

$$p_y = 2.0(0) + 1.0(4.0) = 4.0\text{ kg·m/s}$$

Step 2: Add the masses because they stick together.

$$m_{total} = 2.0 + 1.0 = 3.0\text{ kg}$$

Step 3: Find the final velocity components.

$$v_x = \frac{p_x}{m_{total}} = \frac{6.0}{3.0} = 2.0\text{ m/s}$$

$$v_y = \frac{p_y}{m_{total}} = \frac{4.0}{3.0} \approx 1.33\text{ m/s}$$

So the combined puck moves with a horizontal component of \(2.0\text{ m/s}\) east and a vertical component of about \(1.33\text{ m/s}\) north.

Optional final speed:

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(2.0)^2 + (1.33)^2} \approx 2.4\text{ m/s}$$

Answer: The stuck-together puck moves northeast, with components \(2.0\text{ m/s}\) east and \(1.33\text{ m/s}\) north.

Common mistakes to avoid

• Forgetting direction: Velocity can be positive or negative in 1D.
• Mixing up mass and weight: Momentum uses mass, not weight.
• Assuming kinetic energy is always conserved: Only elastic collisions conserve both momentum and kinetic energy.
• Forgetting to add masses in perfectly inelastic collisions: If objects stick together, use the combined mass after the collision.
• Not separating directions in 2D: Treat \(x\)- and \(y\)-components independently.

Real-world examples

• Car crashes: Investigators use momentum ideas to study collisions.
• Sports: A bat hitting a ball or a hockey puck bouncing off another puck involves momentum transfer.
• Rocket motion: Exhaust gases moving backward help the rocket move forward through momentum conservation.
• Billiards: Pool balls often behave approximately like elastic collisions.

Key formulas

• Momentum of one object: \(p = mv\)
• 1D momentum conservation: \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\)
• Perfectly inelastic collision: \(m_1v_{1i} + m_2v_{2i} = (m_1+m_2)v_f\)
• Kinetic energy: \(KE = \frac{1}{2}mv^2\)
• 2D momentum conservation: conserve momentum separately in \(x\) and \(y\)

Brief Summary

Momentum is the product of mass and velocity, and it is always conserved in an isolated system. In collisions, the total momentum before the collision equals the total momentum after the collision. Elastic collisions conserve both momentum and kinetic energy, while inelastic collisions conserve momentum but not kinetic energy. In perfectly inelastic collisions, objects stick together. In two-dimensional problems, momentum must be conserved separately in each direction.

Rotational Kinematics and Torque help us describe how objects spin, what causes them to start spinning, and what keeps them balanced. Many everyday objects rotate: bicycle wheels, door handles, ceiling fans, seesaws, and even planets. In this lesson, you will connect ideas from straight-line motion to circular motion.

In linear motion, we talk about distance, speed, acceleration, force, and mass. In rotational motion, we use similar ideas, but we describe how things move around an axis. This means we need new quantities such as angle, angular speed, angular acceleration, and torque.

By the end of this lesson, you should be able to:

• Describe angular position, angular velocity, and angular acceleration.
• Relate linear motion to rotational motion.
• Calculate torque using force and distance from a pivot.
• Explain center of mass in simple situations.
• Model rotational equilibrium, when an object is balanced and not turning.

1. Angular Motion: The Rotational Version of Linear Motion

When an object moves in a straight line, we measure how far it travels. When an object rotates, we measure how far it turns using an angle. Angles are often measured in degrees, but in physics we usually use radians.

A full circle is:

This means:

The basic rotational quantities are:

• Angular position \(\theta\): how far something has turned
• Angular velocity \(\omega\): how fast the angle changes
• Angular acceleration \(\alpha\): how fast angular velocity changes

These match the linear quantities:

• Position \(x\)
• Velocity \(v\)
• Acceleration \(a\)

The rotational formulas look very similar to linear formulas:

Here, \(\Delta \theta\) means change in angle, \(\Delta \omega\) means change in angular velocity, and \(\Delta t\) means change in time.

2. Connecting Linear and Rotational Motion

If a point on a spinning object is farther from the center, it travels a longer path in the same amount of time. This is why the outer edge of a merry-go-round moves faster than a point near the middle.

The relationship between linear and rotational motion is:

where:

• \(s\) is the arc length, or distance traveled along the circle
• \(r\) is the distance from the center
• \(\theta\) is the angle in radians

For speed and acceleration:

where:

• \(v\) is linear speed
• \(a_t\) is tangential acceleration, the acceleration along the edge of the circle

This shows that bigger radius means bigger linear speed, even if the angular speed is the same.

3. Rotational Kinematics Equations

If angular acceleration is constant, we can use equations similar to those from linear motion. These are called rotational kinematics equations.

Here:

• \(\omega_0\) is the starting angular velocity
• \(\theta_0\) is the starting angular position
• \(t\) is time

These equations are useful for wheels, fans, gears, and anything rotating with steady angular acceleration.

4. What Is Torque?

In linear motion, a force can change an object's motion. In rotational motion, the turning effect of a force is called torque. Torque tells us how strongly a force tries to rotate an object.

You create more torque when:

• You use a bigger force.
• You apply the force farther from the pivot point.
• You push in a direction that causes turning.

The basic torque formula is:

where:

• \(\tau\) is torque
• \(r\) is the distance from the pivot to where the force is applied
• \(F\) is the force
• \(\theta\) is the angle between the force and the lever arm

If the force is applied perpendicular to the lever arm, then \(\sin\theta = 1\), so the formula becomes:

This is the most common case in basic problems.

The unit of torque is the newton-meter \((\text{N}\cdot\text{m})\).

5. Why Distance from the Pivot Matters

Think about opening a door. It is easier to push near the doorknob than near the hinges. That is because the doorknob is farther from the pivot, so the same force creates more torque.

This is also why long wrenches are useful. A longer wrench gives a larger \(r\), so you can produce more turning effect with the same force.

6. Direction of Torque

Torque can make an object turn clockwise or counterclockwise. In many problems, we choose one direction to be positive and the other to be negative.

• Counterclockwise torque is often positive.
• Clockwise torque is often negative.

You may use the opposite sign rule if you stay consistent throughout the problem.

7. Center of Mass

The center of mass is the balance point of an object or system. It is the point where the object's mass can be thought of as concentrated for many problems.

For a regular, evenly shaped object, the center of mass is usually at the geometric center. For example:

• A uniform ruler has its center of mass at the middle.
• A uniform ball has its center of mass at its center.

For objects with uneven mass or systems made of separate masses, the center of mass shifts toward the heavier side.

Knowing the center of mass helps us understand balance. If the center of mass is directly above the support, the object is less likely to tip. If it moves beyond the support area, the object may rotate and fall.

8. Rotational Equilibrium

An object is in rotational equilibrium when it is not speeding up its rotation. This means the net torque is zero.

If an object is also not moving up, down, left, or right, then the net force is zero too:

For many balancing problems, especially with beams and seesaws, the key idea is:

Total clockwise torque = total counterclockwise torque

When this happens, the object stays balanced and does not rotate.

9. Steps for Solving Torque and Balance Problems

1. Identify the pivot point.
2. List all forces acting on the object.
3. Decide whether each force causes clockwise or counterclockwise rotation.
4. Calculate each torque using \(\tau = rF\) if the force is perpendicular.
5. Set the sum of torques equal to zero if the object is in equilibrium.
6. Solve for the unknown.

Choosing the pivot carefully can make problems easier, because any force acting exactly at the pivot produces zero torque.

Worked Example 1: Angular Velocity

A wheel turns through an angle of \(180^\circ\) in \(2.0\) seconds. What is its angular velocity in radians per second?

Step 1: Convert the angle to radians.

Step 2: Use the formula for angular velocity.

Answer: The angular velocity is about \(1.57\text{ rad/s}\).

Worked Example 2: Relating Linear and Rotational Motion

A bicycle wheel has a radius of \(0.35\text{ m}\). If the wheel spins at an angular speed of \(8.0\text{ rad/s}\), what is the linear speed of a point on the edge?

Use the relationship:

Answer: A point on the edge moves at \(2.8\text{ m/s}\).

Worked Example 3: Calculating Torque

A student pushes on a door with a force of \(15\text{ N}\). The force is applied perpendicular to the door at a distance of \(0.80\text{ m}\) from the hinges. What torque is produced?

Use the torque formula:

Answer: The torque is \(12\text{ N}\cdot\text{m}\).

If the same force were applied halfway to the hinges, the torque would be smaller because the distance \(r\) would be smaller.

Worked Example 4: Rotational Equilibrium on a Seesaw

Two students sit on opposite sides of a seesaw. Student A weighs \(300\text{ N}\) and sits \(2.0\text{ m}\) from the pivot. Student B weighs \(400\text{ N}\). How far from the pivot should Student B sit to balance the seesaw?

For balance, clockwise torque must equal counterclockwise torque.

Substitute the known values:

Answer: Student B should sit \(1.5\text{ m}\) from the pivot.

This example shows that a heavier student can balance a lighter student by sitting closer to the pivot.

10. Common Mistakes to Avoid

• Mixing up degrees and radians: Many rotational formulas require radians.
• Using the wrong distance: Torque uses the distance from the pivot, not the total length of the object unless that is the correct distance.
• Ignoring direction: Clockwise and counterclockwise torques should have opposite signs.
• Forgetting the angle: If the force is not perpendicular, use \(\tau = rF\sin\theta\).
• Confusing force and torque: Force pushes or pulls; torque causes turning.

11. Real-Life Applications

Rotational kinematics and torque are used in many parts of life and engineering:

• Car tires: The wheels rotate while the car moves forward.
• Tools: Wrenches and screwdrivers rely on torque.
• Sports: A bat, racket, or diving motion involves rotation.
• Playground equipment: Merry-go-rounds and seesaws show angular motion and balance.
• Construction: Cranes and beams must stay in rotational equilibrium.

12. Key Idea Connections

It helps to compare linear and rotational motion side by side:

• Position \(x\) matches angular position \(\theta\)
• Velocity \(v\) matches angular velocity \(\omega\)
• Acceleration \(a\) matches angular acceleration \(\alpha\)
• Force \(F\) matches torque \(\tau\)

This comparison makes rotational motion easier to understand, because it builds on ideas you already know from straight-line motion.

Brief Summary

Rotational kinematics describes how objects turn using angular position, angular velocity, and angular acceleration. These quantities connect to linear motion through formulas like \(v = r\omega\) and \(s = r\theta\).

Torque is the turning effect of a force, and it depends on both the size of the force and its distance from the pivot. Rotational equilibrium happens when the net torque is zero, so the object does not rotate. Understanding these ideas helps explain spinning wheels, balanced seesaws, doors opening, and many other everyday situations.

Simple Machines and Mechanical Advantage are important ideas in mechanics because they help us understand how tools make work easier. A simple machine does not reduce the total work that must be done, but it can change how the work is done. For example, it can reduce the force you need to apply by increasing the distance over which you apply that force.

Common simple machines include levers, pulleys, and inclined planes. These machines are used everywhere, from ramps and cranes to scissors and crowbars. To study them, we use the idea of mechanical advantage.

Mechanical advantage tells us how much a machine multiplies force. If a machine allows you to use a smaller input force to overcome a larger output force, then it has a mechanical advantage greater than 1.

In simple terms:

• Input force: the force you apply to the machine
• Output force: the force the machine applies to the object
• Mechanical advantage: how many times the machine increases force

The basic formula for actual mechanical advantage is:

If a machine produces an output force of 200 N when you apply 50 N, then:

This means the machine multiplies your force by 4.

There is also an ideal mechanical advantage, or IMA. This is the mechanical advantage of a perfect machine with no energy lost to friction. IMA depends on the distances involved in the machine.

The general idea is:

This shows an important trade-off: if a machine makes the needed force smaller, you usually have to apply that force over a greater distance.

This trade-off comes from the idea of work. In physics, work is:

where \(F\) is force and \(d\) is distance in the direction of the force.

For an ideal machine, the input work equals the output work:

This means a machine cannot create energy. It only changes the size and direction of forces or the distance over which they act.

In real life, machines are not perfect. Some energy is lost, usually because of friction. Because of this, the actual mechanical advantage is usually less than the ideal mechanical advantage.

We can compare them using efficiency:

A machine with 100% efficiency would be perfect, but real machines always have efficiency less than 100%.

Now let us look at three important simple machines.

1. Levers

A lever is a rigid bar that pivots around a fixed point called the fulcrum. You apply an input force at one point, and the lever applies an output force at another point.

The ideal mechanical advantage of a lever depends on the lengths of the arms:

The input arm is the distance from the fulcrum to where the effort force is applied. The output arm is the distance from the fulcrum to where the load acts.

If the input arm is longer than the output arm, you need less force. This is why a long crowbar makes lifting easier.

2. Pulleys

A pulley uses a rope and wheel to lift loads. A single fixed pulley changes the direction of force but usually does not increase force. A system with multiple supporting rope segments can increase force.

For an ideal pulley system, the ideal mechanical advantage is equal to the number of rope segments supporting the load:

If 4 rope segments support the load, then the ideal mechanical advantage is 4. That means the input force needed is about one-fourth of the load force, ignoring friction.

The trade-off is distance. If the load rises 1 meter, you may need to pull 4 meters of rope.

3. Inclined Planes

An inclined plane is a sloped surface, like a ramp. It helps raise an object to a height using less force than lifting it straight up.

The ideal mechanical advantage of an inclined plane is:

A longer ramp gives a greater mechanical advantage, so less force is needed. However, the object must move a longer distance.

For example, if a ramp is 6 m long and 2 m high:

So, ideally, the ramp reduces the needed force by a factor of 3.

Why force and distance trade places

Simple machines are useful because they let us choose between using more force over a shorter distance or less force over a longer distance. This is not a trick or a violation of physics. It follows the law of conservation of energy.

If you lift a box straight up, you use a larger force over a small distance. If you push the box up a ramp, you use a smaller force, but over a greater distance. The total work is about the same in an ideal situation.

Worked Example 1: Actual Mechanical Advantage

A student uses a lever to lift a rock. The student pushes down with an input force of 80 N, and the lever applies an output force of 240 N to the rock. Find the actual mechanical advantage.

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Solve.

Answer: The actual mechanical advantage is 3. The lever triples the input force.

Worked Example 2: Lever Ideal Mechanical Advantage

A lever has an input arm length of 1.2 m and an output arm length of 0.3 m. Find the ideal mechanical advantage.

Step 1: Use the lever formula.

Step 2: Substitute the values.

Step 3: Solve.

Answer: The ideal mechanical advantage is 4. In an ideal case, the machine can multiply the force by 4.

Worked Example 3: Inclined Plane Force

A ramp is 8 m long and 2 m high. A box has a weight of 400 N. Assume no friction. Find the ideal mechanical advantage and the input force needed to push the box up the ramp.

Step 1: Find IMA.

Step 2: Use mechanical advantage to find input force.

For an ideal machine:

Here, the output force is the weight of the box, 400 N.

Step 3: Solve for \(F_{in}\).

Answer: The ramp has an ideal mechanical advantage of 4, and the needed input force is 100 N.

Worked Example 4: Efficiency of a Pulley System

A pulley system has 3 supporting rope segments, so its ideal mechanical advantage is 3. A worker uses an input force of 120 N to lift a load with an output force of 300 N. Find the actual mechanical advantage and the efficiency.

Step 1: Find AMA.

Step 2: Find efficiency.

Answer: The actual mechanical advantage is 2.5, and the efficiency is about 83.3%.

Important ideas to remember

• A simple machine makes work easier by changing force, distance, or direction.
• Machines do not reduce total work in an ideal case.
• If the input force gets smaller, the input distance usually gets larger.
• AMA is based on real forces.
• IMA is based on distances or machine design.
• Real machines have friction, so efficiency is always less than 100%.

Common mistakes

• Confusing input force with output force
• Forgetting that IMA is for an ideal machine with no friction
• Thinking a machine creates energy
• Using the wrong distance in a lever, such as measuring the whole lever instead of each arm from the fulcrum
• For pulleys, counting the wrong number of rope segments supporting the load

Brief Summary

Simple machines such as levers, pulleys, and inclined planes help people do tasks by changing the size or direction of a force. Their usefulness is measured with mechanical advantage. Actual mechanical advantage compares real output force to input force, while ideal mechanical advantage depends on the machine's shape or distances. These machines make force smaller only by increasing distance, which fits the idea that energy is conserved.

Laws of Thermodynamics help us understand how energy moves and changes in the world around us. They explain why a hot drink cools down, why refrigerators need electricity, and why no machine can be 100% efficient.

In this lesson, you will learn how heat is transferred, what the First Law of Thermodynamics says about energy conservation, and what the Second Law of Thermodynamics says about the natural direction of energy flow.

Thermodynamics is the study of heat, energy, and work. In 10th grade science, you do not need advanced math to understand the main ideas. The most important point is that energy is never just "lost". Instead, it changes form or moves from one place to another.

1. Heat and Temperature

Before learning the laws, it is important to tell the difference between heat and temperature.

• Temperature tells how hot or cold something is.
• Heat is the transfer of thermal energy from a warmer object to a cooler object.

If you touch a metal spoon in a bowl of hot soup, the spoon becomes warm because heat moves from the soup to the spoon.

Heat always moves naturally from higher temperature to lower temperature until the objects reach the same temperature. This is called thermal equilibrium.

2. The Three Main Ways Heat Transfers

Heat can move in three main ways: conduction, convection, and radiation.

Conduction is heat transfer through direct contact. It happens when particles bump into each other and pass along energy.

• A metal pan gets hot on a stove.
• Your hand warms an ice cube when you hold it.
• A spoon left in hot tea becomes hot.

Conduction works especially well in metals because energy moves through them easily.

Convection is heat transfer by the movement of fluids. A fluid can be a liquid or a gas.

• Warm air rises and cool air sinks in a room.
• Water circulates in a boiling pot.
• Ocean currents move warm and cool water around Earth.

Convection happens because warmer fluid usually becomes less dense and rises, while cooler fluid becomes more dense and sinks.

Radiation is heat transfer by electromagnetic waves. Radiation does not need matter, so it can travel through empty space.

• The Sun warms Earth by radiation.
• You feel heat from a campfire even without touching it.
• A toaster heats bread using radiant energy.

3. Systems and Surroundings

In thermodynamics, we often talk about a system and its surroundings.

• The system is the part we are studying.
• The surroundings are everything outside the system.

For example, if we study hot water in a cup, the water may be the system, and the cup, air, and table are the surroundings.

This idea helps us track where energy goes.

4. The First Law of Thermodynamics

The First Law of Thermodynamics states that energy is conserved. Energy cannot be created or destroyed. It can only be transferred or changed from one form to another.

This is the thermodynamics version of the law of conservation of energy.

If a system gains heat, that energy can do one of two things:

• increase the system's stored energy, or
• be used to do work.

A simple way to write the First Law is:

where:

• \(\Delta U\) = change in internal energy of the system
• \(Q\) = heat added to the system
• \(W\) = work done by the system

Internal energy means the energy stored inside a substance because of the motion and arrangement of its particles.

This equation means:

• If heat is added, the system may store more energy.
• If the system does work on its surroundings, some energy leaves the system.

For many 10th grade problems, you can think of it as:

Energy in = increase in stored energy + energy used to do work

Example: In a car engine, fuel releases energy. Some of that energy moves the car, but some becomes heat. The total energy is still conserved.

5. Worked Example 1: Using the First Law

A gas inside a cylinder receives \(500\,J\) of heat. It does \(200\,J\) of work by pushing a piston upward. What is the change in internal energy?

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Calculate.

Answer: The internal energy increases by \(300\,J\).

This makes sense because the system got more energy from heat than it gave away by doing work.

6. The Second Law of Thermodynamics

The Second Law of Thermodynamics explains the natural direction of energy transfer. It says that heat does not flow on its own from a colder object to a hotter object.

Instead, heat naturally flows from hot to cold.

This law also says that in any energy transfer, some energy becomes less useful for doing work. Because of this, no process is perfectly efficient.

The Second Law is closely connected to the idea of entropy.

Entropy is a measure of how spread out energy is or how disordered a system is. A system with higher entropy has energy that is more spread out and less organized.

In simple terms, the Second Law says that the total entropy of a system and its surroundings tends to increase over time.

This does not mean everything always becomes messy in an everyday sense. It means energy naturally spreads out.

Examples of the Second Law:

• Ice melts in a warm room.
• Hot coffee cools down on a table.
• A dropped perfume scent spreads through the air.

These changes happen naturally because energy and particles spread out over time.

7. Why Machines Cannot Be 100% Efficient

An efficient machine changes a large amount of input energy into useful output energy. But because of the Second Law, some energy usually spreads out as waste heat.

For example:

• A light bulb gives off light, but also heat.
• A car engine moves the car, but also becomes very hot.
• A phone charger transfers electrical energy, but may warm up.

That wasted heat means the machine is not 100% efficient.

Efficiency can be written as:

8. Worked Example 2: Efficiency

A machine takes in \(1000\,J\) of energy and produces \(750\,J\) of useful work. What is its efficiency?

Step 1: Use the formula.

Step 2: Substitute the values.

Step 3: Calculate.

Answer: The machine is \(75\%\) efficient.

This means \(25\%\) of the input energy became less useful, often as heat.

9. Heat Transfer and the Laws Together

The three methods of heat transfer connect directly to the laws of thermodynamics.

• Conduction, convection, and radiation explain how thermal energy moves.
• The First Law explains that the total energy is conserved during that movement.
• The Second Law explains why the energy moves in a certain direction, usually from warmer places to cooler ones.

For example, if a hot pan cools on a counter:

• heat leaves the pan by conduction, convection, and radiation,
• energy is conserved as it transfers to the air and nearby objects, and
• the transfer happens naturally from the hotter pan to the cooler surroundings.

10. Worked Example 3: Identifying Heat Transfer

For each situation, identify the main type of heat transfer.

1. A student touches a hot desk lamp and feels heat.
2. Warm air from a heater rises toward the ceiling.
3. Sunlight warms your skin on a cold day.

Solution:

• 1. Conduction — heat transfers by direct contact.
• 2. Convection — warm air moves upward.
• 3. Radiation — energy travels as electromagnetic waves.

11. Worked Example 4: Putting the Ideas Together

A cup of hot chocolate is left on a table. After 20 minutes, it is cooler.

Question A: Did energy disappear?

Answer: No. By the First Law of Thermodynamics, energy was not destroyed. It transferred from the hot chocolate to the cup, air, and table.

Question B: Why did the heat flow out of the drink?

Answer: By the Second Law of Thermodynamics, heat naturally flows from the warmer drink to the cooler surroundings.

Question C: What methods of heat transfer may be involved?

• Conduction to the cup and table
• Convection to the moving air around the cup
• Radiation from the warm surface of the drink and cup

12. Common Misunderstandings

• Misunderstanding: Heat and temperature are the same.
  Correction: Temperature measures how hot something is, while heat is energy being transferred.
• Misunderstanding: Energy is lost when an object cools down.
  Correction: The energy is transferred to the surroundings, not destroyed.
• Misunderstanding: Machines can be perfectly efficient.
  Correction: The Second Law shows that some energy usually becomes waste heat.
• Misunderstanding: Cold flows into an object.
  Correction: It is more accurate to say heat flows out of the warmer object into the cooler one.

13. Key Ideas to Remember

• Heat moves by conduction, convection, and radiation.
• Heat naturally moves from hotter objects to cooler objects.
• The First Law of Thermodynamics says energy is conserved.
• The Second Law of Thermodynamics says energy spreads out and entropy tends to increase.
• No machine is perfectly efficient because some energy becomes less useful, usually as heat.

Brief Summary

The laws of thermodynamics explain how heat and energy behave. The First Law says energy cannot be created or destroyed, only transferred or changed. The Second Law says heat naturally flows from hot to cold and that energy tends to spread out, increasing entropy. Conduction, convection, and radiation are the three main ways heat moves, and these ideas help explain everything from cooling drinks to engines and refrigerators.