Physics help from experts.

Physics Homework Help

Work and Power



Work is done whenever force is applied through a distance. Only the component of the force parallel to the motion does work. As in the diagram below, if a rope is tied to a box and the box is pulled horizontally along a floor, some of the force is pulling upward on the box, and some is pulling in the direction the box is moving.

Work problems sometimes involve force being applied at an angle to the direction of motion, as shown.

The component of force acting in the direction of motion is Fcosø, where ø is the angle between the force and the direction of motion.

Work is found by multiplying distance by the component of force acting in the direction of motion.

The work done by this force is W = Fdcosø, where d is the distance that the object moves while experiencing force, F.

There are three types of sadistic questions often used to test your understanding of work. In these questions, either force, distance, or cosø is zero, and therefore the work done is zero.


1. A man pushes against a car stuck in a snow bank while his date sits nervously behind the steering wheel trying not to make the tires spin. [Other distracting details are usually given, often including numeric information regarding force, weight and so on.] However, the car does not move. How much work did he do on the car?
ANSWER: The man did no work on the car since d=0. He may have burned calories, converting chemical energy into heat, but still, the car did not move.

2. Sally carries a text book under her arm along a horizontal path. [Distracting information is often also given such as the weight or mass of the text book, the distance or more perversely the path taken implying some distance must be calculated and so on.] How much work was done on the text book?
ANSWER: None, since both gravity and the force Sally exerted against gravity are perpendicular to the distance the book moved. (cosø = 0 so W = 0).

3. An asteroid traveling at constant velocity out of reach of gravitational fields [etc.]... How much work is done on the satellite?
ANSWER: None, since F = 0, W = 0.
Another type of question gives the wrong angle.
4. A 200.0 kg load on frictionless coasters is pushed 5 m along a ramp that makes an angle of 20º to the ground. How much work was done on the cart?
ANSWER: Drawing a diagram is a must for any problem involving 2 dimensions or 2 or more forces.

Force diagrams help to understand work problems.

When you do this, the glitch becomes obvious: the angle is not between the force and the distance. The angle between the force and the distance is 70º.
The force applied against gravity is F = mg = (200.0 kg)(9.8 N/kg) = 1960 N
Therefore, W = Fdcosø = 1960(5)(cos70º) = 3352 J

Go to Work and Power Problems


Power is work done per unit of time. The more work done, or the less time taken to do the work, the greater the power.

P =W/t

Usual units of power are J/s which is called "watt." Since watt is a small unit, kilowatt (kW = 1000W) is more commonly used.

Go to Work and Power Problems

Work and Power Problems

(See bottom of page for answers.)

5. A 70 kg cart is pushed for 50 m with a constant velocity upon a 45º frictionless incline. Find the work done on the cart.

6. A 100 kg object is pulled vertically upward 5.0 m by a rope with an acceleration of 1.0 m/s2. Find the work done by the tension force in the rope.

7. A 0.40 kg ball is thrown vertically upward with a speed of 30 m/s. The ball reaches a height of 40 m. What is the energy dissipated due to air friction?

8. If 4 kW of power is dissipated for 30 min, how much energy was involved?

9. A tug pulls on a barge with a constant force of 2.50*106 N, 19° west of north. Another tug pulls on the same barge with the same force but 19° east of north. . What is the total work they done on the barge it is pulled 0.74 km toward the north?

10. A conveyor belt moves for 2.0 m horizontally, then for 2.0 m down an incline angled 10° from the horizontal. A 2.0 kg box is moved by the belt at 0.50 m/s without slipping. At what rate is work being done on the box
a. as the box moves up the 10° incline
b. as the box moves horizontally

11. A 40 kg case is pushed across a floor at a steady speed of 1.5 m/s. When the pushing stops, the case slides a further distance of 1.2 m before coming to rest. Calculate:
a. The frictional force acting on the case when it slides.
b. The work done per second to push the case at a steady speed of 1.5 m/s.

12. A generator harnesses water flowing through a tube 12.5 feet in diameter. About 50% of the kinetic energy of the water is converted to electricity. How fast must the water flow to produce 52 MW of electrical power?

13. A 130 N suitcase is dragged a distance of 250m at a constant speed. A force of 60 N is exerted at an angle 40 degrees above the horizontal.
a. How much work is being done?
b. How much work is done by friction?

(See bottom of page for answers.)

14. How many kilocalories of food energy does a person with a metabolic rate of 97.0 W burn per day?

15. A 1000 kg rocket generating 80 kW of power gains altitude at 3.0 m/s.  What percentage of the engine power is being used to make the rocket climb?

16. How does a person burn calories by just breathing?

17. Calculate the work done by a force acting through a distance, d, if the force changes over the distance according to

F = kx4.

18. Can the normal force on an object ever do work?

19. If 6 million J of energy is consumed in a day, what is this rate of energy consumption in watts?

20. A force of 109 N is applied along a lawnmower's handle, which is 14.7 degrees above the horizontal. If 61.9 W of power was developed for 39 s, what distance is the roller pushed?

21. Suggest a method to measure the power in your legs. Outline the calculations involved.

22. A 1900 kg car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 27 m/s. Going up a hill, the car's engine needs to produce 47 hp more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

23. Find the work done by pushing a mass of 700 kg through a distance of 4.5m along a surface if the coefficient of friction is 0.2. Assume the force is applied horizontally.

24. At a speed of 70 km/h, the average frictional force on a light car is 1050 N. What power must the motor generate to maintain this speed?

25. A winch is used to slide a 350 kg load at constant speed 3.50 m down a 27.0° incline. The rope between the winch and the load is parallel to the incline. The coefficient of kinetic friction between the ramp and the load is 0.40. Calculate the force exerted by the winch, the work done by the winch on the load, the work done by the friction force, the work done by the force of gravity, and the net work done on the load.

26. A 60 kg person developed 70 W of power during a race, dissipating about 0.60 J of mechanical energy per step per kilogram of body mass. If each stride was about 1.5 m, how fast was the person running?

Answers to Work and Power Problems

Selected solutions are printed below.
For solutions to all the problems on this page click here.


Free body diagrams for objects on a slope show force components along the slope or perpendicular to the slope.

Note: Positive is down the inclined plane.

Fy, the component of gravity perpendicular to the inclined plane, is balanced by the normal force, FN.

Fx, the component of gravity doing work on the load is

Fx = mgsinθ = (350)(9.81)sin27.0° = 1559 N

So the work done by gravity is

W = Fd = (1559)(3.50) = 5456 J


FN = Fy = mgcosθ

the force of friction is

μFN = μmgcosθ = (0.40)(350)(9.81)cos27.0° = 1224 N

and the work done by friction is

W = Fd = (-1224)(3.50) = -4283 J

Since the net force on the load is zero (acceleration is zero)

Fx + FF + Fa = 0

The force applied by the winch is

Fa = - Fx - FF = -1559 + 1224 = -335 N

The work done by the winch is

W = Fd = (-335)(3.5) = -1173 J

Since the net force on the load is zero, the net work done on the load is

W = Fd = 0 J.

(Notice the work done by gravity is cancelled by the work done by the winch and friction: 5456 -4283 - 1173 = 0)

For solutions to all the problems on this page click here.

Your satisfaction is our priority.