Physics: Work, Energy, and the Work-Energy Theorem Worksheet

Question 1

A student pushes a box with a constant horizontal force of 50 N. The box moves 6.0 m in the direction of the force. Calculate the work done by the student on the box.

Accepted Answer

The work done is 300 J. Since the force and displacement are in the same direction, W = Fd = 50 N x 6.0 m = 300 J.

Question 2

A rope pulls a sled with a force of 80 N at an angle of 30 degrees above the horizontal. The sled moves 5.0 m horizontally. Calculate the work done by the rope.

Accepted Answer

The work done by the rope is about 346 J. The horizontal component does the work, so W = Fd cos theta = 80 N x 5.0 m x cos 30 degrees = 346 J.

Question 3

A 2.0 kg object speeds up from 3.0 m/s to 7.0 m/s. Use the work-energy theorem to find the net work done on the object.

Accepted Answer

The net work done is 40 J. The work-energy theorem says Wnet = change in kinetic energy = 1/2(2.0 kg)(7.0^2 - 3.0^2) = 40 J.

Question 4

A 1200 kg car slows from 20 m/s to rest during braking. Find the net work done on the car.

Accepted Answer

The net work done on the car is -240,000 J. The car loses kinetic energy, so Wnet = 0 - 1/2(1200 kg)(20 m/s)^2 = -240,000 J.

Question 5

A 15 kg backpack is lifted straight up 2.0 m at constant speed. Calculate the work done by the lifting force.

Accepted Answer

The work done by the lifting force is 294 J. At constant speed, the lifting force equals the weight, so W = mgh = 15 kg x 9.8 m/s^2 x 2.0 m = 294 J.

Question 6

A 5.0 kg box slides 4.0 m across a rough horizontal floor. The coefficient of kinetic friction is 0.20. Calculate the work done by friction.

Accepted Answer

The work done by friction is -39.2 J. The friction force is fk = mu k mg = 0.20 x 5.0 kg x 9.8 m/s^2 = 9.8 N, so W = -fkd = -9.8 N x 4.0 m = -39.2 J.

Question 7

A 60 kg rider starts from rest at the top of a frictionless roller coaster hill and drops 12 m vertically. Find the rider's speed at the bottom.

Accepted Answer

The rider's speed at the bottom is about 15.3 m/s. Gravitational potential energy becomes kinetic energy, so mgh = 1/2mv^2 and v = sqrt(2gh) = sqrt(2 x 9.8 x 12) = 15.3 m/s.

Question 8

A spring with spring constant 200 N/m is compressed by 0.25 m. Calculate the elastic potential energy stored in the spring.

Accepted Answer

The elastic potential energy is 6.25 J. Use Us = 1/2kx^2 = 1/2(200 N/m)(0.25 m)^2 = 6.25 J.

Question 9

A force-displacement graph shows force increasing linearly from 0 N at 0 m to 40 N at 4 m, then staying at 40 N from 4 m to 8 m. Calculate the total work done.

Accepted Answer

The total work done is 240 J. The work is the area under the graph: triangle area = 1/2 x 4 m x 40 N = 80 J, and rectangle area = 4 m x 40 N = 160 J, for a total of 240 J.

Question 10

A 0.15 kg baseball moves at 40 m/s, and a 60 kg runner moves at 2.0 m/s. Compare their kinetic energies.

Accepted Answer

They have the same kinetic energy. The baseball has K = 1/2(0.15 kg)(40 m/s)^2 = 120 J, and the runner has K = 1/2(60 kg)(2.0 m/s)^2 = 120 J.

Question 11

A person pushes a lawn mower with a force of 120 N at an angle of 40 degrees below the horizontal. The mower moves 10 m horizontally. Calculate the work done by the applied force.

Accepted Answer

The work done by the applied force is about 919 J. The horizontal component of the force does the work, so W = Fd cos theta = 120 N x 10 m x cos 40 degrees = 919 J.

Question 12

A 20 kg sled starts from rest. It is pulled horizontally with a 100 N force for 15 m while friction exerts a 25 N force opposite the motion. Find the sled's final speed.

Accepted Answer

The sled's final speed is about 10.6 m/s. The net work is (100 N - 25 N)(15 m) = 1125 J, and Wnet = 1/2mv^2 gives v = sqrt(2Wnet/m) = sqrt(2250/20) = 10.6 m/s.

Question 13

A 0.50 kg ball is thrown straight upward at 18 m/s. Ignore air resistance. Find the work done by gravity from the release point to the highest point.

Accepted Answer

The work done by gravity is -81 J. At the highest point the ball's speed is 0 m/s, so gravity removes all of the initial kinetic energy: Wgravity = 0 - 1/2(0.50 kg)(18 m/s)^2 = -81 J.

Question 14

A 500 kg elevator moves upward 3.0 m while speeding up from rest to 2.0 m/s. Calculate the work done by the cable on the elevator.

Accepted Answer

The work done by the cable is 15,700 J. The net work is the change in kinetic energy, 1/2(500 kg)(2.0 m/s)^2 = 1000 J. Gravity does -mgh = -14,700 J, so the cable must do 1000 J + 14,700 J = 15,700 J.

Question 15

Explain how the sign of work affects an object's kinetic energy. Include one example of positive work and one example of negative work.

Accepted Answer

Positive net work increases an object's kinetic energy, so the object speeds up. For example, a forward push on a cart moving forward does positive work. Negative net work decreases kinetic energy, so the object slows down. For example, friction on a sliding box does negative work.

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Physics: Work, Energy, and the Work-Energy Theorem

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