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CHAPTER 8 Work, Power, and SimpleMachines

Chapter Outline8.1 WORK

8.2 POWER

8.3 REFERENCES

You will recall from the concept on momentum that impulse is defined as force applied times time over which theforce is applied. We found that the same impulse could be delivered by a smaller force applied for a longer periodof time. The concept of work has a similar situation. Work is defined as force applied times the distance the objectmoves due to the applied force. Therefore, the same work can be done by a smaller force applied for a longerdistance.

That is, (5000 N)(2 m) is equal to (50 N)(200 m).

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So, the same amount of work can be done by a 50 N force as is done by a 5000 N force if the 50 N force is appliedover a much greater distance.

We find this fact to be immensely helpful in trying to lift or move large weights. We have designed a group of whatare called “simple machines” (lever, pulley, inclined plane, screw, and wedge) that allow us to move large weightsusing a much smaller force. In the picture of the pulley above, we see that the man by exerting only 100 poundsforce can lift a 400 pound object by using an appropriate pulley system.

In this concept, you will learn all about work and also all about the simple machines that allow us to move weightswith much smaller forces.

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8.1 Work

• Define work.• Identify forces that are doing work.• Given two of the three variables in the equation, W = Fd, calculate the third.

In order for the roller coaster to run down the incline by gravitational attraction, it first must have work done on ittowing it up to the top of the hill. The work done on the coaster towing it to the top of the hill becomes potentialenergy stored in the coaster and that potential energy is converted to kinetic energy as the coaster runs down fromthe top of the hill to the bottom.

Work

The word work has both an everyday meaning and a specific scientific meaning. In the everyday use of the word,work would refer to anything which required a person to make an effort. In physics, however, work is defined as theforce exerted on an object multiplied by the distance the object moves due to that force.

W = Fd

In the scientific definition of the word, if you push against an automobile with a force of 200 N for 3 minutesbut the automobile does not move, then you have done NO work. Multiplying 200 N times 0 meters yields zero

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work. If you are holding an object in your arms, the upward force you are exerting is equal to the object’s weight. Ifyou hold the object until your arms become very tired, you have still done no work because you did not move theobject in the direction of the force. When you lift an object, you exert a force equal to the object’s weight and theobject moves due to that lifting force. If an object weighs 200. N and you lift it 1.50 meters, then your work is,W = Fd = (200. N)(1.50 m) = 300. N m.

One of the units you will see for work is the Newton·meter but since a Newton is also a kilogram·m/s2, then aNewton·meter is also kg·m2/s2. This unit has also been named the Joule (pronounced Jool) in honor of JamesPrescott Joule, a nineteenth century English physicist.

Example Problem: A boy lifts a box of apples that weighs 185 N. The box is lifted a height of 0.800 m. Howmuch work did the boy do?

Solution: W = Fd = (185 N)(0.800 m) = 148 N m = 148 Joules

Work is done only if a force is exerted in the direction of motion. If the force and motion are perpendicular to eachother, no work is done because there is no motion in the direction of the force. If the force is at an angle to themotion, then the component of the force in the direction of the motion is used to determine the work done.

Example Problem: Suppose a 125 N force is applied to a lawnmower handle at an angle of 25° with the groundand the lawnmower moves along the surface of the ground. If the lawnmower moves 56 m, how much work wasdone?

Solution: The solution requires that we determine the component of the force that was in the direction of the motionof the lawnmower. The component of the force that was pushing down on the ground does not contribute to thework done.

Fparallel = (Force)(cos25�) = (125 N)(0.906) = 113 N

W = Fparalleld = (113 N)(56 m) = 630 J

Summary

• In physics, work is defined as the force exerted on an object multiplied by the distance the object moves dueto that force.

• The unit for work is called the joule.• If the force is at an angle to the motion, then the component of the force in the direction of the motion is used

to determine the work done.

Practice

The following video introduces energy and work. Use this resource to answer the questions that follow.

http://wn.com/Work_physics_#/videos

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MEDIAClick image to the left for more content.

1. What definition is given in the video for energy?2. What is the definition given in the video for work?3. What unit is used in the video for work?

The following website contains practice questions with answers on the topic of work.

http://www.sparknotes.com/testprep/books/sat2/physics/chapter7section6.rhtml

Review

1. How much work is done by the force of gravity when a 45 N object falls to the ground from a height of 4.6 m?

2. A workman carries some lumber up a staircase. The workman moves 9.6 m vertically and 22 m horizontally. Ifthe lumber weighs 45 N, how much work was done by the workman?

3. A barge is pulled down a canal by a horse walking beside the canal. If the angle of the rope is 60.0°, the forceexerted is 400. N, and the barge is pulled 100. M, how much work did the horse do?

• work: A force is said to do work when it acts on a body so that there is a displacement of the pointof application, however small, in the direction of the force. Thus a force does work when it results inmovement. The work done by a constant force of magnitude F on a point that moves a distance d in thedirection of the force is the product, W = Fd.

• joule: The SI unit of work or energy, equal to the work done by a force of one Newton when its point ofapplication moves through a distance of one meter in the direction of the force.

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8.2 Power

• Define power.• Given two of the three variables in P = W

t , calculate the third.

Typical Pressurized Water Reactors (PWR) reactors built in the 1970’s produce about 1100 megawatts, whilst thelatest designs range up to around 1500 megawatts. That is 1,500,000,000 Joules/second.

A windmill farm, by comparison, using hundreds of individual windmills produces about 5 megawatts. That is5,000,000 Joules/second (assuming the wind is blowing).

Power

Power is defined as the rate at which work is done or the rate at which energy is transformed.

Power =WorkTime

In SI units, power is measured in Joules per second which is given a special name, the watt, W .

1.00 watt = 1.00 J/s

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Another unit for power that is fairly common is horsepower.

1.00 horsepower = 746 watts

Example Problem: A 70.0 kg man runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5m. Calculate the power output of the man in watts and horsepower.

Solution:

The force exerted must be equal to the weight of the man = mg = (70.0 kg)(9.80 m/s2) = 686 N

W = Fd = (686 N)(4.5 m) = 3090 N m = 3090 J

P = Wt = 3090 J

4.0 s = 770 J/s = 770 W

P = 770 W = 1.03 hp

Since P = Wt and W = Fd, we can use these formulas to derive a formula relating power to the speed of the object

that is produced by the power.

P = Wt = Fd

t = F dt = Fv

The velocity in this formula is the average speed of the object during the time interval.

Example Problem: Calculate the power required of a 1400 kg car if the car climbs a 10° hill at a steady 80. km/h.

Solution: 80. km/h = 22.2 m/s

In 1.00 s, the car would travel 22.2 m on the road surface but the distance traveled upward would be (22.2 m)(sin 10°) =(22.2 m)(0.174) = 3.86 m. The force in the direction of the upward motion is the weight of the car = (1400 kg)(9.80 m/s2)= 13720 N.

W = Fd = (13720 N)(3.86 m) = 53,000 J

Since this work was done in 1.00 second, the power would be 53,000 W.

If we calculated the upward component of the velocity of the car, we would divide the distance traveled in onesecond by one second and get an average vertical speed of 3.86 m/s. So we could use the formula relating power toaverage speed to calculate power.

P = Fv = (13720 N)(3.86 m/s) = 53,000 W

Summary

• Power is defined as the rate at which work is done or the rate at which energy is transformed.• Power = Work

Time• Power = Force⇥velocity

Practice

In the following video, Mr. Edmond sings about work and power. Use this resource to answer the questions thatfollow.

http://www.youtube.com/watch?v=5EsMmdaYClQ

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MEDIAClick image to the left for more content.

1. What units are used for work?2. What units are used for power?

The following website has practice problems on work and power.

http://www.angelfire.com/scifi/dschlott/workpp.html

Review

1. If the circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N, how much workdoes the centripetal force do on the toy when it follows its orbit for one cycle?

2. A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s. How much power does she use?3. (a) Assuming no friction, what is the minimum work needed to push a 1000. kg car 45.0 m up a 12.5°

incline?(b) What power would be needed for the same problem if friction is considered and the coefficient of friction

for the car is 0.30?

• power: The rate at which this work is performed.

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