WORK

We begin with a constant force $F$ that acts along some line that we call the x-axis. By convention $F$ is positive if it acts in the direction of increasing $x$ and negative if it acts in the direction of decreasing $x.$

Suppose now that an object moves along the x-axis from $x=a$ to $x=b$ subject to this constant force $F.$ The work done by $F$ during the displacement is by definition the force times the displacement:
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It is not hard to see that, if $F$ acts in the direction of the motion, then $W>0,$ but, if $F$ acts against the motion, then $W<0.$ Thus, for example, if an object slides off a table and falls to the floor, then the work done by gravity is positive (after all, Earth's gravity points down). But if an object is lifted from the floor and raised to tabletop level, then the work done by gravity is by definition negative.

To repeat, if an object moves from $x=a$ to $x=b$ subject to a constant force $F,$ then the work done by $F$ is the constant value of $F$ times $b-a. $ What is the work done by $F$ if $F$ does not remain constant but instead varies continuously as a function of $x?$ As you would expect, we then define the work done by $F$ as the average value of $f$ times $b-a:$
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Hooke's Law

You can sense a variable force in the action of a steel spring. Stretch a spring within its elastic limit and you feel a pull in the opposite direction. The greater the stretching, the harder the pull of the spring. Compress a spring within its elastic limit and you feel a push against you. The greater the compression, the harder the push. According to Hooke's law (Robert Hooke, 1635 - 1703) the force exerted by the spring can be written
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where $k$ is a positive number, called the spring constant, and $x$ is the displacement from the equilibrium configuration. The minus sign indicates that the spring force always acts in the direction opposite to the direction in which the spring has been deformed (the force always acts so as to restore the spring to its equilibrium state).

Remark. Hooke's law is only an approximation, but it is a good approximation for small displacements. In the problems that follow we assume that the restoring force of the spring is given by Hooke's law.

Problem 1

A spring of natural length $L$ compressed to length $\frac{7}{8}L$ exerts a force $F_{0}.$ Find the work done by the spring in restoring itself to natural length.

Solution

Place the spring on the x-axis so that the equilibrium point falls at the origin. View compression as a move to the left.

Compressed $\frac{1}{8}L$ units to the left, the spring exerts a force $F_{0} $. Thus by Hooke's law
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This tells us that $k=8F_{0}/L.$ Therefore the force law for this spring reads
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To find the work done by this spring in restoring itself to equilibrium, we integrate $F(x)$ from $x=-\frac{1}{8}L$ to $x=0:$
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Problem 2

For the spring in Problem 1, what work must we do to stretch the spring to length $\frac{11}{10}L?$

Solution

To stretch the spring we must counteract the force of the spring. The force exerted by the spring when stretched x units is
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To counter this force we must apply the opposite force
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The work we must do to stretch the spring to length $\frac{11}{10}L$ can be found by integrating $-F(x)$ from $x=0$ to $x=\frac{1}{10}L:$
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If force is measured in pounds and distance in feet, the units of work are footpounds; thus if a force of $500$ pounds pushes a car for $60$ feet, the work done is $30,000$ foot-pounds.~-

Problem 3

Stretched $\frac{1}{3}$foot beyond its natural length, a certain spring exerts a restoring force of $10$ pounds. What work must we do to stretch the spring another foot?

Solution

Place the spring on the x-axis so that the equilibrium point falls at the origin. View stretching as a move to the right. As usual, assume Hooke's law: $F(x)=-kx.$

When the spring is stretched foot, it exerts a force of $-10$ pounds ($10$ pounds to the left). Therefore MATH and $k=30$ (i.e., $30$ pounds per foot). To find the work we must do to stretch the spring another foot, we integrate the opposite force $-F(x)=30x$ from $x=\frac{1}{3}$ to $x=\frac{2}{3}:$
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Pumping Out a Tank

To lift an object we must counteract the force of gravity. Consequently, the work done in lifting an object is given by the formula

work = (weight of the object) x (distance lifted).

If we lift a leaking sand bag or pump out a water tank from above, the calculation of work is more complicated. In the first instance the weight varies during the motion. (There is less sand in the bag as we keep lifting.) In the second instance the distance varies. (Water at the top of the tank does not have to be pumped as far as water at the bottom.) The sand bag problem is left to the exercises. Here we take up the second problem.

6.5 - THE NOTION OF WORK 329

Assume that the liquid is homogeneous and weighs $\sigma $ pounds per cubic foot. Suppose now that this storage tank is pumped out from above until the level of the liquid drops to $b$ feet below the top of the tank. How much work has been done?

We can answer this question by the methods of integral calculus. For each $x\in \lbrack a,b]$ we let

$A(x)$ = cross-sectional area x feet below the top of the tank,

$s(x)$ -- distance that the x-level must be lifted.

We let MATH be an arbitrary partition of $[a,b]$ and focus our attention on the ith subinterval $[x_{i-1},x_{i}]$. Taking $x_{i}^{\ast }$ as an arbitrary point in the ith subinterval, we have

MATH = approximate volume of the ith layer of liquid,

MATH. = approximate weight of this volume, $s(x_{i}^{\ast })$ = approximate distance this weight is to be moved,

and therefore

MATH = approximate work (weight $\times $ distance) required to pump this layer of liquid to the top of the tank.

The work required to pump out all the liquid can be approximated by adding up all these last terms:
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The sums on the right are Riemann sums which, as MATH, converge to give
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We use this formula in the next problem.

Problem 4

A hemispherical water tank of radius $10$ feet is being pumped out. Find the work done in lowering the water level from $2$ feet below the top of the tank to $4$ feet below the top of the tank given that the pump is placed (a) at the top of the tank, (b) $3$ feet above the top of the tank.

Solution. As the weight of water take 62.5 pounds per cubic foot. It is not hard to see that the cross section $x$ feet from the top of the tank is a disc of radius $\sqrt{100-x^{2}}$. Its area is therefore
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For part (a) we have $s(x)=x,$ so that
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For part (b) we have $s(x)=x+3,$ so that
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