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AP Physics C: Mechanics : Understanding Conservation of Energy

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

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Example Questions

Example Question #4 : Work, Energy, And Power

A 0.8kg ball is dropped from rest from a cliff that is 150m high. Use conservation of energy to find the vertical velocity of the ball right before it hits the bottom of the cliff.

Possible Answers:

$32.8\frac{m}{s}$

$40.0\frac{m}{s}$

$92.3\frac{m}{s}$

$54.2\frac{m}{s}$

$73.4\frac{m}{s}$

Correct answer:

$54.2\frac{m}{s}$

Explanation:

The conservation of energy equation is K_1+U_1=K_2+U_2 .

The ball starts from rest so . It starts at a height of 150m, so . When the ball reaches the bottom, height is zero and thus, and $K_2=\frac{1}{2}mv^2$ . The conservation of energy equation can be adjusted below.

U_1=K_2

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

Solve for v.

$v=\sqrt{2gh}$

$v=\sqrt{2(9.8\ \frac{m}{s^2})(150\ m)}$

$v=54.2\ \frac{m}{s}$

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Example Question #1 : Work, Energy, And Power

Starting from rest, a skateboarder travels down a 25^o incline that's 22m long. Using conservation of energy, calculate the skateboarder's speed when he reaches the bottom. Ignore friction.

Possible Answers:

$2.9\frac{m}{s}$

$6.3\frac{m}{s}$

$13.5 \frac{m}{s}$

$32.8\frac{m}{s}$

$21.8\frac{m}{s}$

Correct answer:

$13.5 \frac{m}{s}$

Explanation:

Conservation of energy states that K_1+U_1=K_2+U_2 .

The skateboarder starts from rest; thus, and . At the bottom of the incline, $K_2=\frac{1}{2}mv^2$ and .

U_1=K_2

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

Solve for v.

$v=\sqrt{2gh}$

Using trigonometry, $h=22\sin(25)$ .

$v=\sqrt{2(9.8\ \frac{m}{s^2})(22)(\sin(25))}$

$v=13.5\ \frac{m}{s}$

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Example Question #1 : 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:

$9.9\frac{m}{s}$

$10.8\frac{m}{s}$

$8.9\frac{m}{s}$

$98.0\frac{m}{s}$

$78.4\frac{m}{s}$

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 #1 : Understanding Conservation Of Energy

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:

$9.38\:m/s$

$19.8\:m/s$

$16.0\:m/s$

$11.7\:m/s$

$14.4\:m/s$

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 #3 : 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:

$18.3\:m/s$

$19.5\:m/s$

$21.5\:m/s$

$22.3\:m/s$

$20.5\:m/s$

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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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Understanding Conservation of Energy

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

All AP Physics C: Mechanics Resources

Example Questions

Example Question #4 : Work, Energy, And Power

Example Question #1 : Work, Energy, And Power

Example Question #1 : Energy

Example Question #1 : Understanding Conservation Of Energy

Example Question #3 : Energy

All AP Physics C: Mechanics Resources

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