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CHAPTER 6
CONSERVATION OF ENERGY
So far we have looked at only one form of energy but there are many more. Energy is
a term familiar to all of us but very difficult to explain, even for physicists. It
remains an abstract concept. Energy is a quantity that is associated with the state of
an object. Kinetic energy, for example, is associated with the state of motion of the
object. Despite this abstract definition energy is very useful concept as we will see.
Potential energy
When we throw a ball vertically into the air the ball gradually slows down until it
comes to rest, momentarily, and then falls gradually speeding up. If we consider the
kinetic energy we find that the kinetic energy decreases, becomes zero and then
increases again. If we assume there is no air resistance then the ball will have the
same kinetic energy when it reaches the same height, on the way up and down (since
the speed is the same). The kinetic energy doesn’t simply disappear and then
reappear, instead it changes form from kinetic to potential and then from potential
back to kinetic. Potential energy is energy associated with a system (at least 2
objects) due to the configuration (positions) of that system. In the case of throwing
the ball the system is the ball and the Earth and the configuration has to do with the
height of the ball. Sometimes, however, we simply say the potential energy of the
ball but we must remember that we mean the ball-Earth system.
v=0
K=0
v=0
K=0
v
v0
v=0
K=0
v0
Fig. 6.1 Changes in kinetic energy during motion.
As another example consider a mass at the end of a spring. We can pull the mass,
extending the spring and then let it go. The mass will oscillate, compressing the
spring and then extending again. If the surface is frictionless it will continue this
indefinitely. Again the kinetic energy is increasing and then decreasing (this time
continuously). In this case there is also potential energy associated with the massspring system. Energy is transformed from potential to kinetic and vice versa.
At any time in both the above situations the sum of the potential energy and the
kinetic energy of the system is a constant. This constant is called the mechanical
energy of the system.
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E = U1 + K1 = U2 + K2 = constant
conservation of mechanical energy
where E is the mechanical energy of the system, U1 the potential energy at some time
1 and K1 the kinetic energy at that time. This is called the conservation of
mechanical energy. Another way of writing this is in terms of the changes in kinetic
and potential energies. If the sum is a constant then the change must be zero
K + U = 0
Determining the potential energy
Now we look at calculating the potential energy for different systems. Our equation
above tells us that the change in potential energy is the negative of the change in
kinetic energy
U = K
But our work-kinetic energy theorem says that the change in kinetic energy is equal
to the work done by the net force so that
U = W
If we can calculate the work done by the net force then we can find the change in
potential energy of the system. Note that potential energy has the same units as work
and kinetic energy, the joule (J).
Although the kinetic energy of an object is well defined, it depends on the mass and
speed of the object, the potential energy of a system is not so well defined. This is
because we must select a reference point (configuration) at which the system has zero
potential energy. We can then calculate the potential energy at all other points. We
say that it is only differences in potential energy that matter not the actual values. To
illustrate this let’s take a closer look at our previous examples.
Gravitational potential energy
Consider throwing a ball vertically into the air. We choose the y-axis to be
increasing upwards and take y = 0 at the Earth’s surface. When the ball leaves your
hand the only force acting is the gravitational forcemg (negative since it is pointing
downwards). The change in potential energy between any two points yi and yf is
given by
yf
U = W =   F ( y ) dy
yi
yf
=    mg dy
yi
= mg  y  y f
y
i
= mgyf  mgyi
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If we let yi = 0 then for any other point yf the change in potential energy is simply
mgyf . So if we define the zero of potential energy at y = 0 then the gravitational
potential energy at any other point y is given by
U = mgy
where m is the mass of the object, g the acceleration due to gravity and y the height
above some reference point. The conservation of mechanical energy for a system in
which gravity is the only force acting is then expressed by
E = mgy + ½mv2 = constant
The above is also true for a swinging pendulum even though there is a force due to
the string. This tension force, however, does no work on the pendulum bob and so
we only need to consider the gravitational force. Even when there is sideways
motion, such as in projectile motion gravity is still the only force and so the above is
true.
Elastic potential energy
Consider a mass attached to a spring on a horizontal frictionless surface. The spring
is extended and allowed to oscillate freely. We want to compute the change in
potential energy from an initial position xi to a final position xf. We choose our
reference to be the x-axis and define our zero to be when the spring is in the relaxed
state. The change in potential energy is given by
xf
U = W =   F ( x ) dx
xi
xf
=    kx dx
xi
 
= ½k x 2
= ½k x
2
f
xf
xi
 ½k xi2
If we let xi = 0 then the change in potential energy for any other point xf is given by
½k x 2f . So if we set the zero of potential at x = 0 then the elastic potential energy at
any point x is given by
U = ½kx2
where k is the spring constant and x the displacement from the relaxed state. Again
we could set the zero of potential energy to be anywhere but that would make the
mathematics more complicated. The conservation of mechanical energy for a massspring system becomes
E = ½kx2 + ½mv2 = constant
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The fact that we can choose the zero of potential to be where we like is an advantage
because we should choose it so as to make solving the problem easier.
Q.
A 2.0 kg branch breaks off a tree and falls 5.0 m to the ground below.
What is the branch’s initial potential energy is we take the zero of
potential to be
a) at the ground
b) at the height of a nearby balcony, 3 m above the ground
c) at the height of the branch
What is the change in potential for all the reference systems?
What is the speed of the branch just before it hits the ground?
A.
The gravitational potential energy of the branch (branch-Earth
system) is given by
U = mgy
a) If the ground is the reference point then y = 5.0 m
U = (2.0 kg) (9.8 m.s-2) (5.0 m)
= 98 J
b) If the balcony is 3 m above the ground then the branch is 2.0 m
above it. So
U = (2.0 kg) (9.8 m.s-2) (2.0 m)
= 39 J
c) Since the reference point is at the height of the branch y = 0
U = (2.0 kg) (9.8 m.s-2) (0.0 m)
=0J
5m
3m
The change in potential energy is given by
U = mg(yf - yi)
Since the change in height is the same regardless of the reference
system the change in potential energy will be the same. Writing out
explicitly though we obtain
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a)
b)
c)
U = (2.0 kg) (9.8 m.s-2) ( (0.0 m) - (5.0 m) ) = -98 J
U = (2.0 kg) (9.8 m.s-2) ( (-3.0 m) - (2.0 m) ) = -98 J
U = (2.0 kg) (9.8 m.s-2) ( (-5.0 m) - (0.0 m) ) = -98 J
The conservation of mechanical energy states that
E = U1 + K1 = U2 + K2 = constant
Initially the kinetic energy is zero. If we choose reference system a)
we obtain
98 J + 0 J = 0 J + ½ (2.0 kg) v2
v = 9.9 m.s-1
Q.
The spring of a spring gun is compressed a distance of 3.2 m from its
relaxed state and a ball of mass 12 g is placed in the barrel. With
what speed will the ball leave the barrel once the gun is fired? The
spring constant k is 7.5 N cm-1. Assume no friction and that the gun
is fired horizontally.
A.
The only force we need to consider here is the force exerted by the
spring. There is a gravitational force on the ball but it is
perpendicular to the direction of motion so it does no work (no
change in kinetic energy). So the conservation of mechanical energy
gives
E = U1 + K1 = U2 + K2 = constant
We choose our zero reference for the spring-ball system to be when
the spring is in the relaxed state. Initially the spring is compressed
and so has potential energy but the ball is at rest so has no kinetic
energy. When the ball is ejected the spring-ball system has no
potential energy. All the initial potential energy is converted into
kinetic energy
½ (7.5
N 100 cm

) (0.032 m) + (0 J) = (0 J) + ½ (0.012 kg) v2
cm
1m
v = 8.0 m.s-1
Conservative and nonconservative forces
By now you will have noticed something very peculiar about potential energy and the
force that gives rise to it (like the spring force and gravitational force). When an
object returns to the same position (state) under the action of that force then it has the
same potential energy. Since there is no change in the potential energy the force does
no work over the return journey. Such a force is called conservative.
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A force is conservative if the work it does on an object that
moves through a closed path is zero; otherwise the force is
nonconservative.
Another way of saying the same thing is that if the work done by the force as it
moves an object between two points is the same for all paths connecting those points
then the force is conservative; otherwise the force is nonconservative.
Kinetic frictional force
What would happen to our mass-spring system if there was friction between the mass
and horizontal surface? Eventually, the mass would slow down and stop. The
mechanical energy of the system will gradually decrease. A force such as friction
(and air resistance) is called dissipative. It is also clearly nonconservative. So what
happens to our conservation of mechanical energy? As we have seen energy is
transformed into other forms.
Conservation of energy
In all processes the total energy of an isolated system is conserved. In our previous
example with friction the mechanical energy that is lost is transformed into thermal
energy of the mass and surface. In other words the mass and surface get hot. We are
all familiar with this when we rub our hand across the surface of a table. This
thermal energy is usually written as Eint to signify the internal energy of the system.
Sometimes there are other forms of energy that have to be considered, electrical,
magnetic and so on. As long as we consider all of them we find that the total energy
has not changed. We can write the conservation of energy for an isolated system as
K1 + U1 + E1int + (other forms of energy)1 =
K2 + U2 + E2int + (other forms of energy)2
By isolated system we mean one in which there is no external force. We can always
expand our system to include all the forces acting. If there are forces acting external
to the system then the work done by those forces is equal to the change in the total
energy of the system
W = K + U + int
Work done by friction
When we considered friction acting on our mass-spring system we noted two effects
it produced. It reduced the mechanical energy of the mass-spring system and it
increased the internal energy of the surface and mass. If we consider our system to
be the mass-spring then the frictional force is external and the work done by friction
Wf is given by
Wf = K + U + int
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Surprisingly if the frictional force is given by f and it acts over a displacement d then
the quantity fd is not the work done by the frictional force it is only the loss in
mechanical energy
fd = K + U
This is because friction is a complicated force. The quantity f represents the effect is
has on the motion of the object, not the effect it has on the internal energy of the
object.
Q.
A steel ball whose mass is 5.2 g is fired vertically downward from a
height h1 of 18 m with an initial speed of 14 m.s-1. It travels a further
distance h2 of 21 cm before coming to rest.
What is the change in mechanical energy of the ball?
What is the change in the internal energy of the ball-Earth-sand
system?
What is the average force exerted by the sand on the ball?
A.
The mechanical energy is given by
E = mgy + ½mv2
We set the reference point for potential energy to be the ground
E = Ef - Ei = mg(yf - yi) + ½m( v 2f  vi2 )
= (0.0052 kg) (9.8 m.s-2) (- 0.21 m - 18 m) +
½(0.0052 kg)( (0) - (14 m.s-2)2 )
= -1.4 J
The conservation of energy says that
K1 + U1 + E1int = K2 + U2 + E2int
where we have neglected other forms of energy since they remain the
same.
If we rearrange this equation we find that
E = Eint
So the change in internal energy is 1.4 J. All the mechanical energy
is transformed into internal energy.
For friction we have that
fd = K + U
f (0.21 m) = 1.4 J
f = 6.8 N
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