Roberto’s Notes on Integral Calculus
Chapter 5: Basic applications of integration
Section 11
Work problems
What you need to know already:
 How to use the four step process to set up an
integral.
What you can learn here:
 How to apply this process to the computation
of work done by a force.
 The physic meaning of the concept of work.
We know from physics that if a constant force F is applied to an object and
moves it by a distance s, then the “work” done by the force is defined by the formula
W  Fs .
W   Fi  si
By taking the limit of this sum as the number of pieces increases and their size
decreases, we hope to obtain the exact value through an integral.
Shouldn’t we use d for distance?
Yes, but that would get confused with the same letter used in the differential of
the integral, so we use s instead.
Notice that this formula only applies when the force is constant throughout the
process and all parts of the object are moved by the same amount. But this is
seldom, if ever the case in real applications, as you shall soon see. More
realistically, either:
 the force applied will change continuously (or not) as the process
develops, or
 different portions of the object will be moved by a different amount, with
the change occurring continuously, so that it is impossible to divide the
calculation into finitely many pieces.
Will the slice-approximate-add-limit process that produces integrals allow us to
handle such realistic situations?
You bet! In many cases we can conceptually slice the whole work process into
tiny pieces, for each of which both force and distance are approximately constant:
W  Fi  si
Integral Calculus
In that case we can then add up all such slices:
Chapter 5: Basic applications of integration
But where will the differential dx show up? As part of the force or of the
distance? This will depend on the particular situation, on what portions of the
formula are variable and on what we use as the independent variable.
Therefore, instead of giving you an impossible general formula or procedure, I
will offer you some examples, in the hope that they will clarify how the four step
process can be used in different circumstances.
Example: Work done to stretch a spring
A famous law of physics, called Hooke’s law, states that the force needed to
stretch a spring to a distance s from its natural equilibrium position is given
by F  ks , where k is a constant dependent on the spring. So, to compute
the work needed to stretch a spring by a certain amount, we must consider
that the force needed to stretch the spring is NOT constant, but changes as the
spring gets stretched further and further.
So, let us say that we want to stretch a spring from its current length, a units
from its natural position, to a longer length, b units from such equilibrium.
Section 11: Work problems
Page 1
Example: Work done to lift liquid
When a pump is lifting liquid from a container up to its top, the pump is
performing work only on the liquid in the surface, since any liquid below it is
pushed up by the weight of the liquid on top. But that means that different
parts of the liquid are being moved by different amounts, with the liquid on
top being moved more and more as the container is empties.
a
b



y


y
That being the case, we focus on the work to stretch the spring by such a
small amount that the force may be considered constant:
W  F s  kss
A
S
By adding all the small stretches and taking the limit, we obtain the integral:
b
b
a
a
W   k s ds  k  s ds
To make the problem manageable, we divide the liquid into small horizontal
slice of thickness y (very small) and area A (which may depend on y), so
V  A  y and, if the liquid has
density , each slice has a weight of w  A  y    g .
that each slice has a volume given by
This slice is moved up a distance of y, so that the work needed to lift it is:
For instance, if a spring whose natural length is 8 cm is stretched further to a
length of 12 cm, the work done on it is given by the integral:
4
W  A y    g  y .
By adding up all such small amounts of work and taking the limit, the total
work is given by the integral:
W  k  s ds
b
0
While the work done to stretch it further from here to 15 cm is:
W   g  Aydy
a
7
W  k  s ds
4
Of course this assumes that all these stretches are possible and do not affect
the elasticity of the spring and hence the value of k.
Integral Calculus
Chapter 5: Basic applications of integration
where a and b are the highest and lowest distances that need to be covered.
Remember that distances are from the top, so the lower number corresponds
to the highest point: in applied problems all values must make sense in the
context of the problem!
Section 11: Work problems
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Example: Work done to lift an object of changing mass
Some objects are built in such a way that when you lift them, their mass
changes, either increasing or decreasing. Examples may be leaking buckets
or chains that are being brought up while hanging. In both cases, at any time
the remaining mass of the object changes and therefore the force applied to it
in order to move it changes as well.
Example: Lifting objects to high altitude
As long as we consider lifting objects near the surface of
the Earth, we can assume the acceleration of gravity to be
a constant. But when we plan to lift them to a high
altitude, such as into orbit, we must remember that the
acceleration changes according to Newton’s law of
F G
gravity,
m1m2
, where m1 and m2 are the masses
d2
of the two objects attracting each other, d is the between
their centres of mass and G is the universal constant of
gravitation. For this type of applications we may consider
the two masses to be constant as well, so that the force is
given by
F
k
and the variable here is the distance of
d2
The approach here is to consider a small section of the movement, say at a
height of y and by a distance of y , and figuring out what is the weight of
the system at that point. Usually such weight is a function of y itself, so that
the object from the centre of the Earth. To avoid an awkward use of the letter
d, we shall denote the distance by y.
the work done for such a small section is given by
W 
W  w  y  y .
By adding all the slices and taking the limit, we obtain an integral of the form
For each small segment of the lift, the amount of work is approximated by
k
y , so by adding all such small segments and taking the limit we
y2
obtain the integral:
b
W   w  y  dy ,where a and b, once again, reflect the minimum and
a
b
W
maximum values of the height y considered.

a
k
dy
y2
There are many other possible applications, but the approach is the same. Look
for more in order to better understand such approach and to become an expert in it.
Integral Calculus
Chapter 5: Basic applications of integration
Section 11: Work problems
Page 3
Summary
 To compute the work performed by a force on an object when either the force, or the object or the distance moved change, we can use the four step process to build up the
needed integral.
 There are many variations of this kind of problems and they each need to be analyzed separately. What does not change is the use of the process and the care we must take to
analyze each situation carefully.
Common errors to avoid
 Don’t think that understanding one or two examples is sufficient: these are word problems.
 Need I tell you that this application is very relevant to engineering and most other applied sciences?
Learning questions for Section I 5-11
Review questions:
1. Describe how to use the four-set process to set up an integral representing the
work done by a force on an object when some of the quantities involved are
changing.
2. Describe how to set up an work integral in one particular application of your
choice. The more applications you use, the better it is!
Memory questions:
1. Which formula represents the work done by a force F(x) to move an object over
an interval [a, b]?
4. Which formula represents the work done to lift an object of mass m to a height
h?
2. Which formula describes Hooke’s law?
5. What is the radius of the Earth in km (approximately)?
3. Which formula describes Newton’s law of gravitation?
Integral Calculus
Chapter 5: Basic applications of integration
Section 11: Work problems
Page 4
Theory questions:
1. Why is an integral needed to solve a work problem?
4. Why is an integral need to compute the work needed to lift a flexible object, like
a chain?
2. Why is an integral need to compute the work needed to stretch a spring?
5. In Hooke’s law, what does the independent variable represent?
3. Why is an integral need to compute the work needed to empty a tank from the
6. Does Hooke’s law work for any value of s?
top?
7. Does the variable used for a work integral always represent distance moved?
Application questions:
1. If the work required to stretch a spring to a length of 30 cm is 1.875J and to
stretch it to 40 cm is 16.875J, how long is the spring when at rest?
2. A spring with a natural length of 10cm is stretched to 12cm by a 20g object.
How much work is required to stretch it from equilibrium to 14 cm?
4. Determine the work done to extend a spring from its natural length of 6 inches
to 8 inches, knowing that a 3 lb weight can achieve such stretch.
5. A spring has constant k=12N/m and natural length of 0.5m. Find the work
required to stretch it from a length of 1.5m to a length of 2 m.
3. If the weight of a 5kg object stretches a spring by 3 cm, how much work is done
to stretch the spring by 5 cm from its equilibrium position? What about
stretching from 5 to 10 cm?
6. A tank has the shape of an inverted right circular cone of depth 5m and radius
2m and is filled with water up to 1 metre from the top. What is the amount of
work required to pump all the water to a level 2 metres above the top of the
tank?
8. A tank with the shape of an inverted right square pyramid with base 2m and
height 3 m is full of water. Which integral represents the amount of work done
to empty it by drawing water from the surface and throwing it out over the top?
9. Set up, but do not compute, an integral representing the amount of work needed
7. A cylindrical tank is 10 metres high, 3 metres in diameter, and half full of water.
How much work is done to drain this water out of the tank?
Integral Calculus
Chapter 5: Basic applications of integration
to empty a spherical tank of radius 2 metres from the top if the tank is full of
water to a depth of 1.3 metres.
Section 11: Work problems
Page 5
10. A plugged eaves trough is 5 metres long and has a cross section in the shape of a
right trapezoid, 9 cm deep, 8 cm wide at the top and 5 at the bottom. If it is
filled with water to a depth of 6 cm, how much work needs to be done to empty
it from the top with a floating hose?
11. A conical water tank has diameter 8m and height 8 m with the vertex down.
top if the tank is filled with water to a depth of 5 meters. No need to compute
the integral.
12. A tank with the shape of an inverted right circular cone with h=4m and r=1m is
filled to a depth of 3 metres. How much work is needed to empty the tank from
the top of the tank?
Find an integral representing the work necessary to pump all the water out of the
13. A 9-metre long chain with a mass of 54 kg is hanging from the ceiling of a
warehouse and a hook is attached to it 3 metres from its bottom end. Compute
the work done to pull the hook (with the chain attached to it!) to the ceiling, so
that the chain will be folded into three equal sections.
14. Which integral represents the amount of work done to lift a 20kg bucket from
the bottom of a 10m well by using a rope with a density of 200 g/m?
18. A chain is 10 metres long, weighs 3 kg and is sitting at the edge of the roof of a
tall building, with one end attached to the edge. The other end starts falling
down and pulls the whole chain down until it is hanging by the side of the
building. How much work has been done by gravity to achieve this feat?
19. A 900 kg elevator is suspended by a 50 m cable that weighs 6 kg/m. How much
work is required to raise the elevator from its starting position to a final height
of 20 metres?
15. A 20-meter cable is rolled on the floor and weighs12 kg. Compute the work
needed to lift one end of the cable to a height of 30 metres, leaving the rest of
the cable hanging from it.
16. A pulley is used to obtain draw water from the bottom of a 15-metre well by
means of a bucket. Each time the bucket is filled with 20 litres of water, but as
it rises, it spills water at a rate of 0.1 litres per metre. Compute the total amount
of work done to lift one bucket of water to the top.
20. A 30 metre cable weighing 6 kg is hanging from the top of a 40 metre building.
How much work is required to pull the whole cable to the top of the building?
21. A leaky water bucket is lifted to the top of a 6 meter-tall building by using a
rope and a pulley. The bucket weighs 2kg, it is filled at the beginning with 20L
of water, but by the time it gets to the top (at constant speed) it only holds 14L.
The rope weighs 100g/m. How much work is done to pull up the system?
17. A 30m chain is being pulled up by a crane. If the chain weights 60kg and is
initially on the floor, how much work is done to get it completely stretched
vertically?
22. A rocket weighs 3 tons and is filled with 40 tons of fuel. The fuel is burned at a
rate of 2 tons per 3000 m of climb. Find the work done to boost it to a height of
15000 m.
23. A 1200 kg satellite is to be launched into orbit. How much work is needed to
lift the satellite to a height of 900km?
24. Which integral represents the amount of work done by a rocket to lift a 200kg
GPS satellite to its height of 25000 km?
Integral Calculus
Chapter 5: Basic applications of integration
Section 11: Work problems
Page 6
26. Set up, but do not compute, the integral representing the work needed to move a
25. If the work done by the gas inside the cylinder of an engine to expand the
V2
chamber from volume V1 to V2 is given by the integral
 PdV
and if the gas
V1
fridge magnet 2 units away from the fridge, assuming that the magnetic force
present at a distance of s unit from the magnet is given by
F ( s) 
satisfies the adiabatic equation PV  k , compute the work done by the
engine’s piston to expand the chamber from a volume of 12 cc to a volume of 30
cc if the initial pressure is 60 N/cm2.
1.5
1
52 s  1
2
Templated questions:
When solving any work problem by using an integral, identify:
1. all the quantities that change in the process, thus requiring an integral
2. the units of measurement of the integrand
3. the units of measurement of your final answer
4. the physical quantity represented by the differential
What questions do you have for your instructor?
Integral Calculus
Chapter 5: Basic applications of integration
Section 11: Work problems
Page 7
Integral Calculus
Chapter 5: Basic applications of integration
Section 11: Work problems
Page 8