We begin with a constant force that acts along some line that we call the x-axis. By convention is positive if it acts in the direction of increasing and negative if it acts in the direction of decreasing
Suppose now that an object moves along the x-axis from
subject to this constant force
The work done by
during the displacement is by definition the force times the displacement:
It is not hard to see that, if acts in the direction of the motion, then but, if acts against the motion, then 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
subject to a constant force
then the work done by
is the constant value of
What is the work done by
does not remain constant but instead varies continuously as a function of
As you would expect, we then define the work done by
as the average value of
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
where is a positive number, called the spring constant, and 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.
A spring of natural length compressed to length exerts a force Find the work done by the spring in restoring itself to natural length.
Place the spring on the x-axis so that the equilibrium point falls at the origin. View compression as a move to the left.
units to the left, the spring exerts a force
Thus by Hooke's law
This tells us that Therefore the force law for this spring reads
To find the work done by this spring in restoring itself to equilibrium, we integrate from to
For the spring in Problem 1, what work must we do to stretch the spring to length
To stretch the spring we must counteract the force of the spring. The force
exerted by the spring when stretched x units is
To counter this force we must apply the opposite force
The work we must do to stretch the spring to length can be found by integrating from to
If force is measured in pounds and distance in feet, the units of work are footpounds; thus if a force of pounds pushes a car for feet, the work done is foot-pounds.~-
Stretched foot beyond its natural length, a certain spring exerts a restoring force of pounds. What work must we do to stretch the spring another foot?
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:
When the spring is stretched foot, it exerts a force of
pounds to the left). Therefore
pounds per foot). To find the work we must do to stretch the spring another
foot, we integrate the opposite force
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 pounds per cubic foot. Suppose now that this storage tank is pumped out from above until the level of the liquid drops to 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 we let
= cross-sectional area x feet below the top of the tank,
-- distance that the x-level must be lifted.
We let be an arbitrary partition of and focus our attention on the ith subinterval . Taking as an arbitrary point in the ith subinterval, we have
= approximate volume of the ith layer of liquid,
. = approximate weight of this volume, = approximate distance this weight is to be moved,
= approximate work (weight 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:
The sums on the right are Riemann sums which, as , converge to give
We use this formula in the next problem.
A hemispherical water tank of radius feet is being pumped out. Find the work done in lowering the water level from feet below the top of the tank to feet below the top of the tank given that the pump is placed (a) at the top of the tank, (b) 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
feet from the top of the tank is a disc of radius
Its area is therefore
For part (a) we have
For part (b) we have