Physics Help >> Index to 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. For example, if a rope is tied to a box and the box 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.


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.


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.

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

Problems

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. Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.50*10⁶ N, one 19° west of north and the other 19° east of north, as they pull the tanker 0.74 km toward the north. What is the total work they do on the supertanker?

10. Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50 m/s. At a certain location the conveyor belt moves for 2.0 m horizontally, and finally for 2.0 m down an incline that makes an angle of 10° with the horizontal. Assume that a 2.0 kg box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box

a. as the box moves up the 10° incline

b. as the box moves horizontally

11. A packing case of mass 40 kg is pushed across a rough floor at a steady speed of 1.5 m/s. When the push force is removed, 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 hydroelectric power plant generates 52 MW of electrical power. The falling water (ie, the source of power) passes through an intake tube of 12.5 feet in diameter and turns the shaft of the generator. Assume that 50% of the kinetic energy of the water flowing thru the tube is extracted by the generator. How fast must the water flow through the tube to produce the 52 MW of electrical power?

13. A suitcase with a weight of 130 N is being 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. What work is being done?
b. What work is done by friction?

14. The human body converts internal chemical energy into work and heat at rates of 60 to 125 W (called the basal metabolic rate). This energy comes from food and is usually measured in kilocalories [1 kcal = 4.186 kJ]. (Note that the nutritional `Calorie' listed on packaged food actually equals 1 kilocalorie). How many kilocalories of food energy does a person with a metabolic rate of 97.0 W require per day? (Assume 100 percent efficiency.)

15. Why would two people, both doing the same amount of work, not "feel" the same physically after having done the work.

16. Breathing requires that work be done. How is this work done? Work is cause of which there must be change in energy. What experiences the change in energy?

17. The resistance of a packing material to a sharp object penetrating it is a force proportional to the fourth power of the penetration depth,

F = kx⁴.

a. Calculate the work done to force the sharp object a distance d.

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

19. A couch potato might consume 6 million J of energy in a day. What is the rate of energy consumption in watts?

20. A lawn roller is rolled across a lawn by a force of 109 N along the direction of the handle, which is 14.7 degrees above the horizontal. If the lawn roller develops 61.9 W of power for 39 s, what distance is the roller pushed?

21. How can you measure the power in your legs?

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. A lightweight electric car is powered by 1 ten 12-V batteries.  At a speed of 70 km/h, the average frictional force is 1050 N. What must be the power of the electric motor if the car is to travel at a speed of 70 km/h?

25. A 350 kg Piano slides 3.50 m down a 27.0° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40. Calculate the force exerted by the man, the work done by the man on the piano, the work done by the friction force, the work done by the force of gravity, and the net work done on the piano.

26. While running a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 60 kg person develops a power of 70 W during a race, how fast is the person running? (Assume a running step is 1.5 m long.)

Answers

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

25.

Note: Positive is down the inclined plane.

F_y, the component of gravity perpendicular to the inclined plane, is balanced by the normal force, F_N.

F_x, the component of gravity doing work on the piano is

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

So the work done by gravity is

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

Since

F_N = F_y = mgcosθ

the force of friction is

μF_N = 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 piano is zero (acceleration is zero)

F_x + F_F + F_a = 0

The force applied by the man is

F_a = - F_x - F_F = -1559 + 1224 = -335 N

The work done by the man is

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

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

W = Fd = 0 J.

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