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CHAPTER 5
WORK AND ENERGY
So far we have been looking at problems and analysing them using Newton’s laws of
motion. We now consider another method of solving problems. One which we will
find useful when dealing with objects moving at speeds close to that of light or for
particles that are very small (like the electrons and protons that make up atoms). In
such cases Newton’s laws of motion need to be modified before they are valid. Our
new methods are always valid. They invlove terms that you are already familiar with
but as always physicists give them very precise meanings.
Work: motion in 1-dimension with a constant force
Let us assume that a student pushes an object along a frictionless surface. The force
he applies F is parallel to the surface and the resulting straight line motion. He
pushes the object through a displacement d. We define a quantity work, W, as given
by
W = Fd
where F is the magnitude of F and d the magnitude of the displacement d. We say
that W is the work done by F in moving the object through a displacement d. The
work is done by a force. Note that work is a scalar quantity although force and
displacement are vectors.
F
F
d
d
Now what if the force was not parallel to the motion, as shown in the figure below.
In this case we define the work to be

W = Fd cos 
where  is the angle between the force and the displacement vectors (when placed tail
to tail). We can think of it as the component of the displacement in the direction of
the force or the other way around (they are equivalent). Notice that for our initial
situation  = 0 so our more general equation reduces to the special case.
F
F

d
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d
Sometimes physicists like to write the above equation in a more compact form using
the scalar (or dot) product. It is another way of saying include the cosine of the angle
between the 2 vectors
W = F.d
Our more general equation for work has some more special cases. If the force and
displacement vectors are in opposite directions then  = 180 so that cos  = -1. Since
F and d are always positive this means the work done in such a situation will be
negative. If the force and displacement vectors are perpendicular to each other then 
= 90 so that cos  = 0. In this case there is no work done.
To illustrate these possibilities consider the following situation. You pick up a chair,
carry it with a constant velocity across the room and place it back down again. In the
first part you apply a force F to lift the chair. The force is upwards and so is the
displacement. The force (you) does positive work on the chair. While you are
walking across the room with constant velocity holding the chair you still apply an
upward force F on the chair (otherwise it would fall) but the displacement is
perpendicular to this. The force (you) does no work in this case. Finally as you place
the chair down you still exert an upwards force (as you place the chair down slowly)
but the displacement is downwards so the force does negative work.
Note that throughout the lifting and carrying you must exert a force but that does not
mean you are always doing work, at least not as far as physics is concerned. Your
muscles are straining to support the weight of the chair and they are consuming
energy but not necessarily doing work.
The SI unit for work is the newton-metre since they are the units for force and
displacement. However, work is such a useful quantity that the unit has been given a
special name, the joule (J). It is the work done by a force of 1 N moving through a
displacement of 1 m.
As we have previously seen there can be several forces acting on an object. How do
we calculate the work done in such a case? There are 2 ways. We can first calculate
the net force acting on the object and then use our general equation for work. Or we
can calculate the work done by each of the forces and simply add them up. Either
way we obtain the same answer.
Q.
A weight lifter raises a mass of 250 kg a distance of 2 m. How much work has
he done?
A.
Drawing a free-body diagram
To find the force exerted by the weight lifter we
assume the weight moves at a constant velocity so
the net force must be zero.
F - mg = 0
F = mg
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d
F
The work done is
W = Fd cos
= (250 kg) (9.8 m.s-2) (2 m) cos 0
= 4900 J.
weights
Work done by a variable force.
mg
We now look at the work done by a variable force (one that is not constant). We will
assume that the direction of the force is the same as the displacement (i.e.  = 0) and
then generalise. The force F varies in some arbitrary way that depends on x, F(x), as
indicated in the figure below. We cannot calculate the work done using our previous
equation because the force is not constant. Instead we divide the motion up into
smaller sections, x1, x2, ... xn, each of displacement x. During each of these sections
the force does not change much and we approximate it as constant F ( x ) . The work
done during each small section is W where
W = F ( x ) x
F(x)
F(x)
x
xi
xf
x
xi
x
xf
The total work done is approximately the sum of the work done in each of the
smaller sections.
W =
 W

=
F ( x ) x
We can make this approximation better by making the sections smaller and taking
more of them. In the limit we allow the displacement of the sections to approach
zero, x0 and we obtain the work done by the variable force F(x).
W = lim
x  0

F ( x ) x
This limit is what we define as the integral under the curve F(x). So the work done
by a variable force is given by
W =

xf
xi
F ( x ) dx
where xi and xf are the initial and final positions of the object.
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Variable force in 3-dimensions
For an object acted upon by a force F with components
F = Fx x + Fy y + Fz z
that moves from an initial position (xi, yi, zi) to a final position (xf, yf, zf) the work
done is given by
W =

xf
xi
Fx dx +

yf
yi
Fy dy +

zf
zi
Fz dz
This can also be written in the compact notation as
W =
 F  dr
Work done by a spring
As a practical example of a situation where the force on an object varies (and in fact
with position) we will look at the force exerted by a spring. A spring that is not
being stretched or extended is said to be in its relaxed state. Let us fix one end of the
spring (attach it to a wall or hook) and place an object on the other (free) end. To
measure the displacement of the object we define our origin to be the position when
the spring is in the relaxed state. Furthermore, we consider only 1-dimension, x,
along the axis of the spring. When we stretch the spring the there is a force in the
opposite direction to the displacement trying to restore the spring to its relaxed state.
When we compress the spring the force is again opposite to the displacement trying
to restore the spring to the relaxed state. This force due to the spring is called the
restoring force. We usually define the x-axis such that when the spring is extended
x is positive and when the spring is compressed x is negative.
x=0
F=0
x negative
F positive
x positive
F negative
F
F
x
x
x
0
0
0
Now it turns out that to a good approximation for most springs the force F is
proportional to the displacement x from the relaxed state. We write the force as
F = -kx
and this is called Hooke’s law. The constant k is termed the spring constant and is
a property of each individual spring. It is a measure of how stiff (difficult to stretch)
the spring is. The SI units for the spring constant are newton per metre (N.m-1). The
minus sign in Hooke’s law indicates that the force is in the opposite direction to the
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displacement. Although the above is a vector equation you will often see it written in
terms of the components as
F = -kx
Hooke’s law gives us the explicit dependence of the force on the displacement so that
it is possible to calculate the work done by the spring using our integral method. Let
us assume that we move the above object (attached to a spring) from an initial
position xi to a final position xf. The work done by the spring is
W =

xf
xi
F ( x ) dx =
 
= (½k) x

xf
xi
xf
 kx dx = k  x dx
xi
2 xf
xi
= (½k)( x  xi2 )
2
f
Note that since F(x) is the force exerted by the spring it is the work done by the
spring. The work done by us is the negative of this (we exert an opposite force).
If the initial position of the object (on the spring) is at xi = 0 then the work done in
moving it to a final position of xf is
W = ½k x 2f
Q.
An object is attached to a spring with spring constant 500 N.m-1. If
the spring is stretched by 10 cm from the relaxed state how much
work does the spring do? If it is stretched another 10 cm how much
work is done?
A.
Work done by the spring is given by
W = (½k)( x 2f  xi2 )
where xi = 0 m and xf = 10 cm = 0.10 m. Substituting we obtain
W = (½ (500 N.m-1)) ( (0.10 m)2  (0 m)2 )
= 2.5 J
To move a further 10 cm we have xi = 10 cm = 0.10 m and xf = 20 cm
= 0.20 m
W = (½ (500 N.m-1)) ( (0.20 m)2  (0.10 m)2 )
= 7.5 J
{The spring does negative work in each case but the force stretching
the spring does positive work.}
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Kinetic energy
We have looked at calculating the work done by a force acting on an object but what
about the object. Consider the following. A boy picks up a ball and then throws it.
We neglect the initial lifting and focus on the actual throw. During the throwing
motion the boy exerts a force on the ball. This force is in the same direction as the
motion so it does positive work. When the ball eventually leaves the boy’s hand it is
moving with a speed v. What is it about the state of the ball, after the throw, that tells
you work has been done on it?
There is a property of any object (including the ball) called the kinetic energy. The
kinetic energy is defined to be
K = ½mv2
where m is the mass of the object and v the speed. The SI units for kinetic energy are
the same as for work, the joule (J). The kinetic energy of an object is always positive
(or zero) it can never be negative.
We can relate the kinetic energy of an object to the total work done on that object (or
the work done by the net force) by
W = K f - KI
where Ki and Kf are the initial and final kinetic energies of the object. This is called
the work-kinetic energy theorem and says that the total work done on an object is
equal to the change in kinetic energy of that object. So when the boy does work on
the ball he also increases the kinetic energy of the ball.
Q.
A pot plant falls from a ledge and drops 10 m to the ground below.
How fast was it falling just before hitting the ground?
A.
We can solve this problem either by using our equations for free-fall
or the work-kinetic energy theorem. Let’s us the later.
The only force acting during free-fall is due to gravity
F = mg
Since this is constant and in the same direction as motion
W = mg x
The initial speed is 0 so Ki = 0.
The work-kinetic energy theorem gives
mg x = ½m( v 2f  vi2 )
vf =
=
2gx
2 (9.8 m.s-2 ) (10 m) = 14 m.s-2
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Power
Sometimes we are not interested in the work done by a force but the rate at which the
work is done. For example the top speed of a car is determined by the rate at which
the motor can do work. This quantity is called the power and the average power is
defined to be
P =
W
t
where W is the work done in a time t. The instantaneous power is given by
P =
dW
dt
The SI units for power are the joule per second (J.s-1) but it is also used very often
and so has a special name, the watt (W).
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