Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 12: Applications Of The Integral – Section 12.8: Work

12.8
Work

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1. Work

When a force acts on an object and moves the object for some distance in the direction of the force, we say that the
force has done work on the object.

Suppose a force F has done work on an object to move it along the x-axis from point x = a to point x = b > a. If the
force is constant, say F = K where K is a constant, throughout the interval [a, b], then the work W done by F is defined
to be the product of the force F and the distance s = b – a. So we have W = Fs. See Fig. 1.1. The value of W is equal to
that of the area of the colored region.

Assume that the force F isn't constant but depends on the position x of the object; more precisely, that F changes
continuously as x increases from a to b, in such a way that F is a continuous function of x, ie, F = F(x) is a continuous

Fig. 1.1 For a constant force, work = force x distance.

Fig. 1.2 Total work:

Fig. 1.3 Total work:

Note that the value of W is equal to the area of the colored region in Fig. 1.3. Note too that Eq. [1.1] is also applicable if
F = K is constant:

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2. Practical Treatment

In Part 1 above, we used the rigorous theoretical partition-and-Riemann-sum argument to arrive at Eq. [1.1]. We're now
going to present the practical treatment of how to arrive at the integral of Eq. [1.1]. The practical treatment is the same
as the theoretical one if the force F is constant: W = Fs. Following is the practical treatment for the situation where F
isn't constant.

Suppose that a force F moves an object on the x-axis from x = a to x = b and that F changes continuously with position
x of the object, ie, F = F(x) is a continuous function. At any point x in [a, b], the work element done by F over a short
distance dx from x to x + dx is dW = F(x) dx. So, adding these little pieces of work from x = a to x = b, we obtain the
work W done by F in moving the object from x = a to x = b:

That's it, the practical treatment. Observe that it uses differential notation. Its justification is the theoretical treatment,
discussed in Part 1. The practical treatment is used in solving problems. It's also closer to that of typical physics or
engineering texts than is the theoretical one.

Example 2.1

Hooke's law states that the force required to stretch or compress a spring to x units longer or shorter than its natural
length is, for sufficiently small values of x, proportional to x. That is, F(x) = kx for some constant k, called the spring
constant. Suppose that a force of 10 kg is required to stretch a spring by 5 cm. Find the work done in stretching the
spring that far.

Solution
By Hooke's law, F(x) = kx. Now 10 = F(5) = 5k, so k = 2. Thus F(x) = 2x. The work element done to stretch the
spring by dx is dW = F(x) dx = 2x dx. As a consequence the work done in stretching the spring by 5 cm is:

EOS

Since work = force x distance for a constant force or work = sum of products force x differential distance for general
force, if force is in kg and distance in cm then work is in kg . cm.

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3. Slicing The Object

Example 3.1

Water fills a cylindrical tank of radius 2 m and height 6 m. All of the water is pumped out of the tank over the top edge of
the tank. The density of water is 1000 kg/m³. How much work is done in emptying the tank?

Solution

Fig. 3.1 A slice of water at point y has thickness dy.

EOS

In Eq. [1.1] or [2.1]:

where x is distance, to obtain the work W.

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Problems & Solutions

1. Two electrons a distance s apart repel each other with a force inversely proportional to the square of the distance, ie,
F(s) = k/s² for some constant k. Suppose one electron remains fixed at the origin of the x-axis. Find the work done by
the force of repulsion in moving the other electron from the point 1 to the point 3 on the x-axis.

Solution

The work done in moving the other electron a distance dx from x to x + dx is dW = F(x) dx = (k/x²) dx. So the work
done in moving it from the point 1 to the point 3 is:

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2. Let R be the radius of the Earth in km. The force of gravity F that attracts an object of mass m kg located at a height
of h km above the surface of the Earth is given by:

where K is a constant and F is in kg. Determine the work that must be done against gravity to raise the object from
the surface of the Earth to:
a. A height of H km.
b. An infinite height.

Solution

a. Let h be an arbitrary height in km in [0, H] above the surface of the Earth. The work done to raise the object by dh km
from h km to (h + dh) km is:

Notes

i) Since work = force x distance for a constant force or work = sum of products force x differential distance for general
force, if force is in kg and distance in km then work is in kg . km.

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3. A bucket is full of water. The water is leaking at a constant rate of 1.2 kg/min. The bucket is raised vertically from
ground level at a constant speed of 10 m/min by a winch. When it starts up it weighs 21 kg. How much work must be
done by the winch to raise it to a height of 30 m?

Solution

Let h be an arbitrary height in m in [0, 30]. The time elapsed from when the bucket starts up to when it reaches the
height of h m is (h/10) min. From when the bucket starts up to (h/10) min afterward it's lost water by (1.2)(h/10) =
(0.12)h kg, so its total weight then is (21 – (0.12)h) kg, thus the force applied on it by the winch then is (21 – (0.12)h) kg.
The work done on it by the winch to raise it by dh m from h m to (h + dh) m is dW = (21 – (0.12)h) dh kg . m. The work
that must be done by the winch to raise it to a height of 30 m is:

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4. A long flexible chain lies coiled on the floor. It weighs 3 kg/m. Find the work done in lifting one end of the chain
vertically 1.5 m off the floor.

Solution 1 – Force Changing Continuously With Position

When that end of the chain is at a height of h m above the floor, the lifted part of the chain has a mass of 3h kg. So the
force needed to lift that end to that height is F(h) = 3h kg. The work done in lifting it dh m from h to h + dh is dW =
F(h) dh = 3h dh kg . m. Thus the work done in lifting that end of the chain vertically 1.5 m off the floor is:

Note

As F(h) = 3h, the force F applied to the lifted end of the chain to lift it changes continuously with the height or position h
of that end.

Solution 2 – Slicing

Consider the situation where the lifted end of the chain is at the height of 1.5 m. For any height h where h is in [0, 1.5]
consider a slice of the lifted part of the chain at height h with thickness dh. The mass of this slice is 3 dh kg. So the force
needed to lift the slice to the height of h m is 3 dh kg. The work done by this force on the slice is dW = (3 dh)h =
3h dh kg . m. Adding up the works done on all of the slices we get the required work:

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5. A tank is in the shape of a right circular cone with height 5 m and top radius 2 m. It contains water up to the height of
4 m. The density of water is 1,000 kg/m³. How much work must be done to pump all of the water out of the tank over
the top edge of the tank?

Solution

Note

We integrate from 0 to the initial height of the water, which is 4, not from 0 to the height of the tank, which is 5, unless
they’re the same. If we integrate from 0 to 5, then for any initial height of the water, 4 m or 1 m or 3 m or any value in
[0, 5], the outcome of integration would be the same, which can’t be correct. So integration from 0 to 5 can’t be correct.
The reason for integration from 0 to the initial height of the water say h_w, 4 in this case, is that [0, h_w] is the domain of
the variable h (thus the slice is obtained only on [0, h_w], and we add up the work elements on the slices that are obtained
there). Now does the height of the tank say h_t, 5 in this case, have a role to play? Yeah it does, as seen in the solution.

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