§ 7.5 - WorkWork Done by a Constant Force

Work occurs when a force moves an object some distance in the direction of theforce. The amount of work for a constant force is just the product of the force andthe distance moved.

Work = constant force * distance moved in the direction of the force

If we lift an object that weighs 50 pounds, then we have to exert a force of 50pounds (actually a hair more at the beginning to start it moving), and if we move it 4feet, then we have done W = (50 pounds)(4 feet) = 200 foot-pounds of work.

W = F ⋅ D

Note that the work done is just the area under the force-distance curve.

Work Done by a Variable Force

In most cases of interest, the force is not constant. It varies over the distancemoved. We could break the area under the force-distance curve into a sequence ofnarrow vertical rectangles, and treat the force as constant for each little rectangle,but we’ve already learned that in the limit of an infinite number of infinitesimallynarrow rectangles, we can just integrate the force function over the distancemoved to find the area, and hence the work done.

Example 1 - Doing work by compressing (or stretching) a spring

The force required to compress (or stretch) a spring increases in direct proportionto the distance we compress it. In other words, F = kx, where x is the distance infeet the spring is compressed and k is a measure of how difficult it is to compressthe spring (measured in pounds per foot). The force-distance curve for ahypothetical spring with k = 2 pounds/foot is shown below.

W = F(x) dx∫b

a

The work done by compressing the spring a distance of 4 feet is

W = F(x) dx = 2x dx = = 16 f oot − pounds∫b

a ∫4

0[ ]x2 4

0

Note that it took a maximum force of 8 pounds, but the average force was only 4pounds, and an average force of 4 pounds over a distance of 4 feet is 16 foot-pounds of work.

Example 2 - Doing work by lifting a space module into orbit agains gravity

Newton’s law of gravity says that the force of the Earth’s gravity (or any other mass)decreases as the square of the distance from its center.

where C is a constant that depends on the mass of the Earth and the mass of theobject being lifted. For this problem assume the weight of a space module is 15metric tons and that the radius of the Earth is 4000 miles. Then

The work to lift a space module from the Earth’s surface (4000 miles from theEarth’s center) to a height of 800 miles (4800 miles from the Earth’s center) isshown by the red area under the force-distance curve

F = Cx2

C = F = 15 tons ⋅ (4000 miles = 240, 000, 000 = 2.4 × ton − milex2 )2 108 s2

Example 3 - Doing work by lifting oil out of a spherical tank

A spherical tank of radius 8 feet is half full of oil that weighs 50 pounds per cubicfoot. Find the work required to pump oil out through a hole in the top of the tank.

W = F(x) dx = dx = 2.4 ×∫b

a ∫4800

4000

2.4 × 108

x2 108[− ]1x

4800

4000

= 2.4 × (− + ) = 10000 ton − miles108 14800

14000

If we imagine an infinitesimally thin “disk” of oil at a depth y below the middle ofthe sphere, then that disk of oil has to be raised a distance 8-y (note that y isnegative), as y goes from 0 to –8. The volume of the disk of oil is its area (πx ) timesits thickness dy, and the force that has to be exerted to lift it is the density of oil (50pounds per cubic foot) times the disk volume.

For the sphere, x and y are related by

2

Work = F(y)(8 − y) dy = π (8 − y) dy∫−8

0 ∫−8

0x2

x = −82 y2‾ ‾‾‾‾‾‾√

Work = 50 π(64 − )(8 − y) dy = 50 π(512 − 64y − 8 + ) dy∫−8

0y2 ∫

8

0y2 y3

Work = 50π = 589, 782 f oot − pounds−8

Example 4 - Doing work by lifting a chain

A 20-foot chain weighing 5 pounds per foot is lying coiled on the ground. Howmuch work is required to raise one end of the chain to a height of 20 feet so that itis fully extended?

If we think of each link in the chain being of length dy, then the weight of that link is5 dy, and each link in the chain is being raised to a height y, where y varies from 0to 20 feet.

As a quick check, we can say the whole chain weighs 5 pounds/foot times 20 feet= 100 pounds, and the average height is 10 feet, so the work is 100 pounds times10 feet = 1000 foot-pounds.

Example 5 - Work done by an expanding gas

Work = 50π = 589, 782 f oot − pounds[512y − 32 − + ]y2 83

y3 14

y4−8

0

Work = 5y dy = 5 = 1000 f oot − pounds∫20

0 [ ]y2

2

20

0

A quantity of gas with an initial volume of 1 cubic foot and a pressure of 500pounds per square foot expands by pushing a piston to increase the volume to 2cubic feet. Find the work done by the gas. Note that as the volume of a gasincreases, its pressure decreases inversely (assuming temperature is constant). Forexample, if we double the volume of a gas, its pressure will only be half as great. Ifwe double the pressure on a gas, its volume will be reduced to half the earliervolume. This is summarized by the relation P = k/V, where k is a constant thatdepends on the amount of gas and the temperature. Also, note that pressure isforce per unit area, so if we have a cylinder of uniform cross section, for example,as the gas expands its moving a piston some distance x. The distance x times thecross-sectional area of the cylinder is the change in volume, then the work done is

If P = 500 pounds per square foot when V = 1 cubic foot, then k = 500 foot-pounds.

Work = P dV = dV∫V2

V1∫

V2

V1

kV

Work = dV = 500 ln|V = 500 ln 2 = 346.6 f oot − pounds∫2

1

500V

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