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Principles of Work Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction to Principles of Work

One of the basic principles of physics is that work performed is force times distance: If you apply force F pounds in moving an object d feet, then the work is

W = F · d foot-pounds.

The problem becomes more interesting (than simple arithmetic) if the force is varying from point to point. We now consider some problems of that type.

Examples

Example 1

A weight is pushed in the plane from x = 0 to x = 10. Because of a prevailing wind, the force that must be applied at point x is F ( x ) = 3x2x + 10 foot-pounds. What is the total work performed?

Solution 1

Following the way that we usually do things in calculus, we break the problem up into pieces. In moving the object from position x to position x + Δx , the distance moved is Δx feet and the force applied is about F ( x ) = 3x2x + 10. See Fig. 8.23 . Thus work performed in that little bit of the move is w ( x ) = (3x2x + 10) · Δx. The aggregate of the work is obtained by summation. In this instance, the integral is the appropriate device:

Applications of the Integral 8.3 Work

 

Applications of the Integral 8.3 Work

Fig. 8.23

Example 2

A man is carrying a 100 lb sack of sand up a 20-foot ladder at the rate of 5 feet per minute. The sack has a hole in it and sand leaks out continuously at a rate of 4 lb per minute. How much work does the man do in carrying the sack?

Solution 2

It takes four minutes for the man to climb the ladder. At time t , the sack has 100 − 4 t pounds of sand in it. From time t to time t + Δ t , the man moves 5 · Δ t feet up the ladder. He therefore performs about w ( t ) = (100 − 4 t ) · 5 Δ t foot-pounds of work. See Fig. 8.24 . The total work is then the integral

Applications of the Integral 8.3 Work

Applications of the Integral 8.3 Work

Fig. 8.24

You Try It: A man drags a 100 pound weight from x = 0 to x = 300. He resists a wind which at position x applies a force of magnitude F ( x ) = x 3 + x + 40. How much work does he perform?

Examples

Example 3

According to Hooke’s Law, the amount of force exerted by a spring is proportional to the distance of its displacement from the rest position. The constant of proportionality is called the Hooke’s constant . A certain spring exerts a force of 10 pounds when stretched 1/2 foot beyond its rest state. What is the work performed in stretching the spring from rest to 1/3 foot beyond its rest length?

Applications of the Integral 8.3 Work

Fig. 8.25

Solution 3

Let the x -variable denote the position of the right end of the spring (Fig. 8.25), with x = 0 the rest position. The left end of the spring is pinned down. Imagine that the spring is being stretched to the right. We know that the force exerted by the spring has the form

F ( x ) = kx,

with k a negative constant (since the spring will pull to the left). Also F (0.5) = −10. It follows that k = −20, so that

F ( x ) = −20 x.

Now the work done in moving the spring from position x to position x + Δ x will be about (20 x ) · Δ x (the sign is + since we will pull the spring to the right— against the pull of the spring). Thus the total work done in stretching the right end of the spring from x = 0 to x = 1/3 is

Applications of the Integral 8.3 Work

Example 4

Imagine that a water tank in the shape of a hemisphere of radius 10 feet is being pumped out ( Fig. 8.26 ). Find the work done in lowering the water level from 1 foot from the top of the tank to 3 feet from the top of the tank.

Applications of the Integral 8.3 Work

Fig. 8.26

Solution 4

A glance at Fig. 8.27 shows that the horizontal planar slice of the tank, at the level x feet below the top, is a disk of radius Applications of the Integral 8.3 Work.

Applications of the Integral 8.3 Work

Fig. 8.27

This disk therefore has area A ( x ) = π · (100 − x2). Thus a slice at that level of thickness Δ x will have volume

V ( x ) = π · (100 − x 2 ) · Δ x

and (assuming that water weights 62.4 pounds per cubic foot) weight equal to

w ( x ) = 62.4π · (100 − x 2 ) · Δ x .

Thus the work in raising this slice to the top of the tank (where it can then be dumped) is

W ( x ) = [62.4π · (100 − x 2 ) · Δ x ] · x foot-pounds.

We calculate the total work by adding all these elements together using an integral. The result is

Applications of the Integral 8.3 Work

You Try It: A spring has Hooke’s constant 5. How much work is performed in stretching the spring half a foot from its rest position?

Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.

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