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AP Physics C Electricity : Energy

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #1 : Energy

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:

$40.0\frac{m}{s}$

$73.4\frac{m}{s}$

$92.3\frac{m}{s}$

$32.8\frac{m}{s}$

$54.2\frac{m}{s}$

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 #2 : Energy

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:

$21.8\frac{m}{s}$

$32.8\frac{m}{s}$

$13.5 \frac{m}{s}$

$2.9\frac{m}{s}$

$6.3\frac{m}{s}$

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 #3 : 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:

$98.0\frac{m}{s}$

$78.4\frac{m}{s}$

$8.9\frac{m}{s}$

$10.8\frac{m}{s}$

$9.9\frac{m}{s}$

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 #1 : 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:

$16.0\:m/s$

$9.38\:m/s$

$14.4\:m/s$

$19.8\:m/s$

$11.7\:m/s$

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 #2 : 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:

$20.5\:m/s$

$18.3\:m/s$

$21.5\:m/s$

$19.5\:m/s$

$22.3\:m/s$

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 #3 : 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:

$51.65 \frac{m}{s}$

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

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

$0 \frac{m}{s}$

$54.25 \frac{m}{s}$

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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Example Question #4 : Energy

An object has a mass of 5kg and has a position described by the given function:

x(t)=2x^2-5x+12

What is the object's kinetic energy after two seconds?

Possible Answers:

7.5J

27.5J

22.5J

45.2J

Correct answer:

22.5J

Explanation:

Kinetic energy is defined by the equation:

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

Taking the derivative of the position function allows us to obtain the velocity function:

x'(t)=v(t)=4x-5

We can now determine the velocity after two seconds:

v(2)=4(2)-5=3

Now that we know our velocity, we can solve for the kinetic energy.

$KE=\frac{1}{2}(5kg)(3\frac{m}{s})^2=22.5J$

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Example Question #5 : Energy

An object starts from rest and accelerates at a rate of $2.5\frac{m}{s^2}$ . If the object has a mass of 10kg, what is its kinetic energy after three seconds?

Possible Answers:

75J

226.5J

281.25J

562.64J

Correct answer:

281.25J

Explanation:

Kinetic energy is given by the equation:

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

We can find the velocity using the given acceleration and time:

$v=v_o+a\Delta t$

$v=0\frac{m}{s}+(2.5\frac{m}{s^2})(3s)=7.5\frac{m}{s}$

Use this velocity to find the kinetic energy after three seconds:

$KE=\frac{1}{2}(10kg)(7.5\frac{m}{s})^2=281.25J$

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Example Question #6 : Energy

A 120kg box has a kinetic energy of 2300J. What is its velocity?

Possible Answers:

$40.1\frac{m}{s}$

$19.4\frac{m}{s}$

$6.2\frac{m}{s}$

$38.3\frac{m}{s}$

$7.2\frac{m}{s}$

Correct answer:

$6.2\frac{m}{s}$

Explanation:

The formula for kinetic energy of an object is:

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

The problem gives us the mass and the kinetic energy, and asks for the velocity, so we can rearrange the equation:

$v=\sqrt{\frac{2KE}{m}}$

Use our given values for kinetic energy and mass to solve:

$v=\sqrt{\frac{2(2300J)}{120kg}}=6.2\frac{m}{s}$

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Example Question #7 : Energy

Calculate how much potential energy a vertical standing spring gains if someone puts a 4kg box on top of it. Take the spring constant to be 120N/s.

Possible Answers:

8.0J

8.9 J

6.5 J

12.2 J

3.0J

Correct answer:

6.5 J

Explanation:

First, find out how much the spring compresses when the 4kg box is put on top of it. To find this, consider what forces are acting on the box when it is resting on the spring. The forces acting on it are the upward spring force, , and the downward gravitational force,; thus, the net force acting on the box is given by the equation below.

$F_{net}=F_s-F_g=-kx-mg=0$

We have chosen the upward direction to be positive and the downward direction to be negative. The net force is equal to zero because the box is at rest. Solve for x.

$x=-\frac{mg}{k}$

$x=-\frac{(4\ kg)(-9.8\ \frac{m}{s^2})}{120\ \frac{N}{m}}$

$x=0.33\ \text{m}$

Then use this value to find the potential energy of the spring.

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

$U=\frac{1}{2}(120\ \frac{N}{m})(0.33\ m)^2$

$U=6.5\ J$

So when the box is placed on top of the spring, the spring gains a potential energy of 6.5J.

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Energy

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #1 : Energy

Example Question #2 : Energy

Example Question #3 : Energy

Example Question #1 : Energy

Example Question #2 : Energy

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All AP Physics C Electricity Resources

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