SESSION 7

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.

Fig. 7.1 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 mass-spring 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.

E = U1 + K1 = U2 + K2 = constantconservation 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 yiand yf is given by

U = W = 

= 

= mg

= mgyf mgyi

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

U = W = 

= 

= ½k

= ½k ½k

If we let xi = 0 then the change in potential energy for any other point xf is given by ½k. 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 mass-spring system becomes

E = ½kx2 + ½mv2 = constant

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)

= 0 J

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

a)U = (2.0 kg) (9.8 m.s-2) ( (0.0 m) - (5.0 m) ) = -98 J

b)U = (2.0 kg) (9.8 m.s-2) ( (-3.0 m) - (2.0 m) ) = -98 J

c)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 ) (0.032 m) + (0 J) = (0 J) + ½ (0.012 kg) v2

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.

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

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()

= (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 themechanical 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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