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AP Physics C: Mechanics : Work, Energy, and Power

Study concepts, example questions & explanations for AP Physics C: Mechanics

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Example Questions

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Example Question #1 : Understanding Conservation Of Energy

A $\small 5.0kg$ bowling ball is dropped from $\small 5.0m$ above the ground. What will its velocity be when it is $\small 1.0m$ above the ground?

Possible Answers:

$78.4\frac{m}{s}$

$10.8\frac{m}{s}$

$98.0\frac{m}{s}$

$9.9\frac{m}{s}$

$8.9\frac{m}{s}$

Correct answer:

$8.9\frac{m}{s}$

Explanation:

Relevant equations:

K_i+U_i=K_f+U_f

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

U = mgh

Determine initial kinetic and potential energies when the ball is dropped.

K_i = 0

$U_i = (5kg)(9.8\frac{m}{s^2})(5m)=245J$

Determine final kinetic and potential energies, when the ball has fallen to $\small 1.0m$ above the ground.

$K_f=\frac{1}{2}(5kg)v_f^2$

$U_f = (5kg)(9.8\frac{m}{s^2})(1m)=49J$

Use conservation of energy to equate initial and final energy sums.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

Solve for the final velocity.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

$v_f^2=\frac{245J-49J}{\frac{1}{2}(5kg)}=78.4\frac{m^2}{s^2}$

$v_f=8.85 \frac{m}{s} \approx 8.9\frac{m}{s}$

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Example Question #11 : Work, Energy, And Power

A solid metal object with mass of $25\: kg$ is dropped from rest at the surface of a lake that is $20\:m$ deep. The water exerts a drag force on the object as it sinks. If the total work done by the drag force is - $3800\: J$ , what is the speed of the object when it hits the sand at the bottom of the lake?

Possible Answers:

$14.4\:m/s$

$9.38\:m/s$

$16.0\:m/s$

$11.7\:m/s$

$19.8\:m/s$

Correct answer:

$9.38\:m/s$

Explanation:

This is a conservation of energy problem. First we have to find the work done by gravity. This can be found using:

$W= F\cdot d, F = mg \\ W = 25\:kg \cdot 9.8\:m/s^2 \cdot 20\:m = 4900\:J$

It is given to us that the work done by the drag force is $-3800\:J$ which means that work is done in the opposite direction. We take the net work by adding the two works together we get, $1100\:J$ of net work done on the block.

Since this is a conservation of energy problem, we set the net work equal to the kinetic energy equation:

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

is the mass of the block and we are trying to solve for .

$v = \sqrt{\frac{2 \cdot 1100}{25}} = 9.38\:m/s$

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Example Question #1 : Energy

If a roller coaster car is traveling at $15\: m/s$ when it is $24\: m$ above the ground, how fast is it going when $10\: m$ above the ground?

Possible Answers:

$21.5\:m/s$

$18.3\:m/s$

$19.5\:m/s$

$22.3\:m/s$

$20.5\:m/s$

Correct answer:

$22.3\:m/s$

Explanation:

This is a classic conservation of energy problem. We know that potential energy and kinetic energy both have to conserve. So we use the following equation:

$\frac{1}{2}mv_o^2+mgh_1 = \frac{1}{2}mv_f^2 + mgh_2$

What this equations says is, the sum of kinetic and potential energy is same at varying heights and velocities.

We can simplify this equation by cancelling out all the m terms.

$\frac{1}{2}v_o^2+gh_1 = \frac{1}{2}v_f^2 + gh_2$

We know all the terms except for v_f , which is the final speed we are trying to solve for h_1 , which is $24\:m$ , h_2 is $10\:m$ and v_o is $15\:m/s$ .

If we plug in all the numbers and solve for v_f , we get $22.3\:m/s$ .

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Example Question #1 : Interpreting Work, Energy, And Power Diagrams

An object with a mass of 25kg is moving at $25\frac{m}{s}$ in a straight line on a fricitonless surface. After a force of 250N acting in the direction of its motion is applied to it for 13.75m , what is the object's speed in meters per second?

Possible Answers:

$30\frac{m}{s}$

$31\frac{m}{s}$

$28\frac{m}{s}$

$32\frac{m}{s}$

$29\frac{m}{s}$

Correct answer:

$30\frac{m}{s}$

Explanation:

Begin by using the following equation relating the initial and final kinetic energy and the work done on the object:

KE_i + W = KE_f

Then, plug in the given variables and solve for the final speed.

$\frac{1}{2}mv_i^2 + Fd = \frac{1}{2}mv_f^2$

$\frac{1}{2}(25kg)(25\frac{m}{s})^2 + (250N)(13.75m) = \frac{1}{2}(25kg)v_f^2$

Simplify terms.

$7812.5 + 3437.5 = \frac{25}{2}v_f^2$

Isolate the final velocity and solve.

$v_f = \sqrt{11250*\frac{2}{25}}$

$v_f = \sqrt{900}$

$v_f = 30\frac{m}{s}$

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Example Question #131 : Mechanics Exam

A 30kg projectile is launched straight upwards at an initial velocity of $32.65\frac{m}{s}$ . What is the maximum height that this projectile reaches in meters?

Round to the nearest meter, and assume the projectile encounters no air resistance.

Possible Answers:

56m

53m

55m

54m

52m

Correct answer:

54m

Explanation:

You can use the motion equation and find the maximum, but it may be faster to use energy equations. Set the initial kinetic energy equal to the gravitational potential energy at the maximum height and solve for the height.

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

Mass cancels.

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

Isolate the height and solve.

$h = \frac{v^2}{2g} = \frac{(32.65\frac{m}{s})^2}{2(9.8\frac{m}{s^2})} = 54.3889m$

Round to 54m .

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Example Question #11 : Work, Energy, And Power

A set of cars on a roller coaster with a combined mass of 1000kg is at the top of its initial hill and will drop 45m down the hill before the track starts to rise again. What will the coaster's speed be at the moment the track starts to rise again?

Round to the nearest meter per second. You may also assume the track does not create friction.

Possible Answers:

$29\frac{m}{s}$

$30\frac{m}{s}$

$28\frac{m}{s}$

$32\frac{m}{s}$

$31\frac{m}{s}$

Correct answer:

$30\frac{m}{s}$

Explanation:

Remember that gravitational potential energy is not affected by the path downward (or upward)—whether it is straight, curved, or winding—only by how big the drop is. Once that is taken into account, you can simply set the initial gravitational potential energy and final kinetic energy equal to each other as if the coaster were falling straight down and solve for the final velocity.

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

The mass cancels.

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

Isolate the velocity and solve.

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8\frac{m}{s^2})(45m)}$

$v = 29.7\frac{m}{s}\approx30\frac{m}{s}$

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Example Question #1 : Interpreting Work, Energy, And Power Diagrams

A plane weighing 1500kg dives 40m with its engine off before it goes into a circular pattern with a radius of 200m while maintaining its speed at the end of its dive. How much centripetal force, in Newtons, is acting on the plane?

Possible Answers:

6000N

5760N

5640N

6120N

5880N

Correct answer:

5880N

Explanation:

First, find the gravitational potential energy of the drop. Then, set it equal to the kinetic energy at the end of the drop and solve for the velocity.

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

The mass cancels.

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

Isolate the velocity and solve.

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8\frac{m}{s^2})(40m)}$

$v = 28\frac{m}{s}$

This gives you the last term you need to solve for the centripetal force.

$F_c = \frac{mv^2}{r}$

$F_c = \frac{(1500kg) (28\frac{m}{s})^2}{200m} = 5880N$

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Example Question #1 : Interpreting Work, Energy, And Power Diagrams

If the maximum speed of an object attached to the end of a elastic has a 1:1 ratio (a meter per second for each meter) with how much the elastic is stretched or compressed from its starting position, which of the following is true?

Possible Answers:

The spring constant has a 4:1 ratio with the mass of the object in kilograms

The spring constant has a 1:1 ratio with the mass of the object in kilograms

None of these statements are true

The spring constant has a 2:1 ratio with the mass of the object in kilograms

The spring constant has a 1:2 ratio with the mass of the object in kilograms

Correct answer:

The spring constant has a 1:1 ratio with the mass of the object in kilograms

Explanation:

Set the elastic potential and kinetic energy equations equal to each other:

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

You are given the fact that in this case, x^2 = v^2 . This allows you to simplify the equality.

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

$\frac{1}{2}k = \frac{1}{2}m$

k = m

This shows us that there is a 1:1 ratio between the spring constant of the elastic and the mass of the object.

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Example Question #1 : Interpreting Work, Energy, And Power Diagrams

A 10kg pumpkin is being launched out of an air cannon. For safety reasons, the pumpkin cannot be more than 100m off the ground during flight, and this particular cannon always launches pumpkins at $80\frac{m}{s}$ meters per second—any more power and the pumpkin could be blasted apart; any less and the pumpkin may not leave the launch tube.

What is the maximum possible angle of launch in degrees?

Round to the nearest whole degree.

Possible Answers:

$30^\circ$

$27^\circ$

$39^\circ$

$34^\circ$

$36^\circ$

Correct answer:

$34^\circ$

Explanation:

We know the maximum height of the pumpkin, which tells us the maximum energy of the launch. Calculate the final gravitational potential energy.

$PE_g = mgh = (10kg ) (9.8\frac{m}{s^2}) (100m) = 9800J$

Now set this value equal to a kinetic energy equation that uses the vertical component of the velocity at launch.

$9800 = \frac{1}{2}mv_y^2$

$v_y = \sqrt{\frac{9800J * 2}{10kg}}$

$v_y = 44.27\frac{m}{s}$

Use the vertical velocity component to determine the launch angle.

$\theta = \sin^{-1}(\frac{v_y}{v})$

$\theta= \sin^{-1}(\frac{44.27\frac{m}{s}}{80\frac{m}{s}}) = 33.60^\circ\approx34^\circ$

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Example Question #1 : Kinetic Energy

A train car with a mass of 2400 kg starts from rest at the top of a 150 meter-high hill. What will its velocity be when it reaches the bottom of the hill, assuming that the bottom of the hill is the reference level.

Possible Answers:

$0 \frac{m}{s}$

$51.65 \frac{m}{s}$

$54.25 \frac{m}{s}$

$51.65 \frac{m}{s^2}$

$55.0 \frac{m}{s^2}$

Correct answer:

$54.25 \frac{m}{s}$

Explanation:

The law of conservation of energy states:

PE_i + KE_i = PE_f + KE_f

If the car starts at rest, then the initial kinetic energy = 0 J.

If the car ends at the reference height, the final potential energy = 0 J.

Subsituting these values, the equation becomes:

PE_i + 0 J = 0 J + KE_f

The initial potential energy can be determined by:

$PE_i = mgh_i \\ PE_i = 2400 kg \cdot 9.81 \frac{m}{s^2} \cdot 150 m\\ PE_i = 3531600 \frac{kg \cdot m^2}{s^2}\\ PE_i = 3531600 J$

The final kinetic energy equation is:

$KE_f = \frac{1}{2}mv_f^2\\ KE_f = \frac{1}{2}(2400 kg)v_f^2$

Substituting the initial potential energy and final kinetic energy into our modified conservation of energy equation, we get:

$3531600 J = \frac{1}{2}(2400 kg)v_f^2\\ 3531600 \frac{kg\cdot m^2}{s^2} = 1200 kg \cdot v_f^2\\ \frac{3531600 \frac{kg\cdot m^2}{s^2}}{1200 kg}=v_f^2\\ 2943 \frac{m^2}{s^2}=v_f^2\\ \sqrt{2943 \frac{m^2}{s^2}}=\sqrt{v_f^2}\\ 54.25 \frac{m}{s}=v_f$

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AP Physics C: Mechanics : Work, Energy, and Power

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Understanding Conservation Of Energy

Example Question #11 : Work, Energy, And Power

Example Question #1 : Energy

Example Question #1 : Interpreting Work, Energy, And Power Diagrams

Example Question #131 : Mechanics Exam

Example Question #11 : Work, Energy, And Power

Example Question #1 : Interpreting Work, Energy, And Power Diagrams

Example Question #1 : Interpreting Work, Energy, And Power Diagrams

Example Question #1 : Interpreting Work, Energy, And Power Diagrams

Example Question #1 : Kinetic Energy

All AP Physics C: Mechanics Resources

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