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AP Physics C Electricity : AP Physics C

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Example Questions

Example Question #3 : Work, Energy, And Power

An object is pushed across a rough surface with a force of 53N. The rough surface exerts a frictional force of 3.47N on the object. If the object is pushed 7.9m, how much work is done on the object?

Possible Answers:

1452.88 J

446.11 J

391.29 J

398.63J

326.47J

Correct answer:

391.29 J

Explanation:

The defintion of work is:

$W=F_{net}\cdot d$

The net force on this object is:

$F_{net}=F_{app}-F_{friction}$

We can calculate this term using the given values:

$F_{net}=53N-3.47N=49.53N$

The distance is given. Substituting these values:

W=49.53 N (7.9m)=391.29 J

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Example Question #3 : Work, Energy, And Power

ONe of the forces that act on a particle as it underjoes a dispalcement of

2i + 3j -1k m is given by $F = 8i - 2j +3k\: N$ .

How much work is done by that force on the particle?

Possible Answers:

$1\:J$

$9\:J$

$5\:J$

$7\:J$

$3\:J$

Correct answer:

$7\:J$

Explanation:

The equation of work is given by $W = F\cdot d$ , where is force and is the distance. Both the work and the force are given to us, but in vector form. In this case, we have to take the dot product of the force and distance.

When taking dot product, keep these rule in mind

$i \cdot i = 1, j \cdot j = 1, k \cdot k =1$

Every other dot proudct is equal to 0.

$(8i - 2j +3k) \cdot (2i + 3j - 1k) = \\((8\cdot2)+(-2\cdot3)+(3\cdot-1) ) = 7\:J$

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Example Question #121 : Mechanics Exam

When riding your bicycle, you have a maximum power output of 500W. As you approach a hill, you shift into first gear, applying a torque of $30\ N\cdot m$ to the gears. Assuming you produce 50% of your maximum power output, what is the angular velocity of the gears, in radians per second?

Possible Answers:

$16.66\frac{rad}{s}$

$10.0\frac{rad}{s}$

$8.33\frac{rad}{s}$

$15.5\frac{rad}{s}$

Correct answer:

$8.33\frac{rad}{s}$

Explanation:

Power is determined by calculating the work output per unit time. In this case, power will be the product of torque and angular velocity:

$P=\tau*\omega$

We are given values for our torque and our power output, allowing us to solve for the angualr velocity. Keep in mind that the power output is going to be 50% of the maximum.

$250W=(30N\cdot m)(\omega)$

$\omega=8.33\frac{rad}{s}$

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Example Question #2 : Work, Energy, And Power

A crane lifts a crate with a mass of 50kg. The crate is raised at a constant velocity for ten seconds and and moves a vertical distance upwards of 20m. What power is being supplied to the crane during this time?

Possible Answers:

100W

980W

830W

730W

Correct answer:

980W

Explanation:

For this problem we can calculate the power as the product of force and velocity:

P=Fv

First, we need to find the velocity:

$v=\frac{\Delta x}{\Delta t}=\frac{20m}{10s}=2\frac{m}{s}$

Our force will be equal to the weight of the crate:

$F=mg=(50kg)(9.8\frac{m}{s})=490N$

Now, we can solve for power:

$P=Fv=(490N)(2\frac{m}{s})=980W$

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Example Question #1 : Work, Power, And Diagrams

A person is moving boxes up the stairs in their new home. They have two identical boxes, with same the size and mass. The first box is easy to carry up the stairs. When moving the second box, the person is more tired and moves more slowly. Which statement accurately describes the work and power between the two trials?

Possible Answers:

Work decreases; power stay the same

Work increases; power decreases

Work remains the same; power decreases

Work remains the same; power increases

Work increases; power increases

Correct answer:

Work remains the same; power decreases

Explanation:

Because the boxes are the same mass and are moving the same distance, the work done will remain the same between the two instances. Work does not depend on time:

W=Fd

However, when carrying the second box, the person moves more slowly. The overall time increases, which leads to a decrease in power.

$P=\frac{W}{t}$

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Example Question #2 : Work, Energy, And Power

What power is required to lift a 25kg box 5.76m off the ground in fifteen seconds?

Possible Answers:

9.6 W

3.77 W

1412.64 W

16.35 W

94.18W

Correct answer:

94.18W

Explanation:

The definition of power is:

$P=\frac{W}{t}$

The work done on the box to lift it is required in order to overcome the force of gravity. If we can find the force of gravity on the object, we can calculate the net force. The force of gravity on any object near the Earth's surface is:

F_g=mg

The definition of work is:

$W=F_{net}d$

We can substitute the force of gravity for the net force, resulting in the equation:

W=mgd

Substituting this into our power equation, we find:

$P=\frac{mgd}{t}$

Plugging in our given values and constants, we find:

$P=\frac{(25 kg) (9.81 \frac{m}{s^2})(5.76m)}{15 s}$

$P=\frac{(25)(9.81)(5.76)}{15}\frac{kgm^2}{s^3}$

P=87.36 W

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Example Question #3 : 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:

$40.0\frac{m}{s}$

$73.4\frac{m}{s}$

$32.8\frac{m}{s}$

$54.2\frac{m}{s}$

$92.3\frac{m}{s}$

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 #4 : 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:

$6.3\frac{m}{s}$

$32.8\frac{m}{s}$

$2.9\frac{m}{s}$

$21.8\frac{m}{s}$

$13.5 \frac{m}{s}$

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 #11 : Work, Energy, And Power

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:

$9.9\frac{m}{s}$

$8.9\frac{m}{s}$

$78.4\frac{m}{s}$

$10.8\frac{m}{s}$

$98.0\frac{m}{s}$

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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All AP Physics C Electricity Resources

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AP Physics C Electricity : AP Physics C

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

All AP Physics C Electricity Resources

Example Questions

Example Question #3 : Work, Energy, And Power

Example Question #3 : Work, Energy, And Power

Example Question #121 : Mechanics Exam

Example Question #2 : Work, Energy, And Power

Example Question #1 : Work, Power, And Diagrams

Example Question #2 : Work, Energy, And Power

Example Question #3 : Work, Energy, And Power

Example Question #4 : Work, Energy, And Power

Example Question #11 : Work, Energy, And Power

Example Question #1 : Understanding Conservation Of Energy

All AP Physics C Electricity Resources

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