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3.2Conservation of energy
Potential energy is a property of a system, not of an individual particle. Potential energy, like kinetic energy, is a scalar quantity. Why is the quantity EP = mgh important enough or useful enough that we give it a special name? To answer this question, consider a body that falls freely under gravity along a vertical line, Figure 3.19.

Figure 3.19: A body falling vertically increases in speed as the height decreases.

When the body is at height h1 from the ground its speed is v1 and when at height h2 its speed is v2. We know from kinematics that
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v22=v21+2g(h1âh2)
which we may rewrite as
12v22+gh2=12v21+gh1
Multiplying through by the mass m of the falling object we get
12mv22+mgh2=12mv21+mgh1
This is an interesting result: each side of the equation is the sum of the kinetic energy of the falling mass and the quantity we called gravitational potential energy. The equation suggests that this sum, EK + EP, stays the same as the mass falls. So if we call this sum the total mechanical energy of the system, ET, we are led to a law of conservation of total mechanical energy. Of course, we showed that ET stays the same for the simple case of a mass falling freely, but this can be demonstrated for many other situations as well. This is why the concept of gravitational potential energy is useful.
So, for the system consisting of the particle and the earth, the total mechanical energy is
ET=EK+EP=12mv2+mgh
Under what conditions is ET conserved? It turns out that this happens when friction and resistance forces are absent and no other forces act on the system from outside the system. Clearly, a car travelling on a level road will come to rest when the engine is turned off. Total mechanical energy is not conserved here because of the work done by the resistance forces opposing the motion; total mechanical energy decreases. Similarly, if the driver of the car steps on the gas pedal, the car will accelerate, increasing the kinetic and hence total mechanical energy. This is because an external force has acted on the car (between the ground and the tyres), and again this external force did work, this time increasing the total mechanical energy. So we are led to this conclusion:
KEY POINT
If no work is done on the system from outside, the system is called isolated and in that case the total mechanical energy of the system does not change. We say that the total mechanical energy of the system is conserved.
If our system includes springs the total mechanical energy will be
ET=EK+EP+EH=12mv2+mgh+12kx2
In the absence of external forces doing work this total mechanical energy will be conserved.
(It is important to realise that weight and spring tension are not considered external forces because their effects are included in the potential energy in the total mechanical energy of the system.)
CHECK YOURSELF 3
A ball starting from rest will descend the same vertical height along the paths shown in Figure 3.20. The length of the path increases from left to right.
In which case will the final speed be greatest?

Figure 3.20: For check yourself question 3.

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WORKED EXAMPLE 3.10
Determine the minimum speed of the mass in Figure 3.21 at the initial point such that the mass makes it over the barrier of height h = 0.80 m.

Figure 3.21: For worked example 3.10.

Answer
To make it over the barrier the mass must be able to reach the highest point. Any speed it has at the top will mean it can carry on to the other side. Therefore, at the very least, we must be able to get the ball to the highest point with zero speed.
With zero speed at the top, the total energy at the top of the barrier is ET = mgh.
The total energy at the starting position is ET=12mv2.
Equating the initial and final total energies,
12mv2=mgh
âv=2ghââââ
Thus, the initial speed must be bigger than v=2ghââââ=2Ã9.81Ã0.80âââââââââââââ=3.96â4.0Â mÂ sâ1.
Note that if the initial speed v of the mass is larger than 4.0 m sâ1, then when the mass makes it to the original level on the other side of the barrier, its speed will be v. Note also that the mass plays no role in this calculation.
WORKED EXAMPLE 3.11
A ball rolls off a 1.0 m high table with a speed of 4.0 m sâ1, as shown in Figure 3.22. Calculate the speed as the ball strikes the floor.

Figure 3.22: For worked example 3.11.

Answer
The total energy of the mass is conserved. As it leaves the table with speed u it has total energy given by EinitialÂ =12mu2+mgh, and as it lands with speed v the total energy is EfinalÂ =12mv2 (v is the speed we are looking for).
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Equating the two energies gives
12mv2=12mu2+mgh
âÂ v2v2=u2+2gh=16+19.62=35.62
âvâ6.0Â mÂ sâ1
Notice that if the ball was projected at an angle to the horizontal the result would be the same. Can you see why?
We mentioned earlier that the height can be measured from any horizontal surface, not just the surface of the earth. Let us see how this works in this example. Suppose first that we measure heights from a level that is 3 m below the surface of the earth. The total energy of the ball on the table would then be 12mu2+mgÃ4. When the ball lands the total energy would be 12mv2+mgÃ3. Equating the two we get
12mu2+mgÃ4=12mv2+mgÃ3
â12mu2+mgÃ1=12mv2
âv2=u2+2gÃ1=16+19.62=35.62
just like before. Similarly, suppose we measure heights from the level of the table. The total energy at the table is now 12mv2+0. The total energy on the ground is 12mv2+mgÃ(â1) since the ground is 1 m below the table. Equating again,
12mu2+0=12mv2+mgÃ(â1)
â12mu2+mgÃ1=12mv2
âv2=u2+2gÃ1=16+19.62=35.62
again just like before. So, in using gravitational potential energy we can measure heights from any horizontal surface that is convenient.
WORKED EXAMPLE 3.12
Two identical balls are launched from a table with the same speed u (Figure 3.23). One ball is thrown vertically up and the other vertically down. The height of the table from the floor is h. Predict which of the two balls will hit the floor with the greater speed.

Figure 3.23: For worked example 3.12.

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Answer
At launch both balls have the same kinetic energy and the same potential energy. When they hit the floor their energy will be only kinetic. Hence the speeds will be identical and equal to v, where
12mv2âv2âv=12mu2+mgh=u2+2gh=u2+2ghââââââââ
WORKED EXAMPLE 3.13
A pendulum of length 1.0 m is released from rest with the string at an angle of 10Â° to the vertical. Find the speed of the mass on the end of the pendulum when it passes through its lowest position.
Answer
We measure heights from the lowest point of the pendulum (Figure 3.24). The total energy at that point is just kinetic, EK=12mv2, where v is the unknown speed.

Figure 3.24: For worked example 3.13.

At the initial point, the total energy is just potential, EP = mgÎh, where Îh is the vertical difference in height between the two positions. From the diagram,
Îh=1.0â1.0cos10Â°
Îh=0.015Â m
Equating the expressions for the total energy at the lowest point and at the start,
12mv2=mgÎh
v=2gÎhââââââ
v=0.54Â mÂ sâ1
Note how the mass has dropped out of the problem. (At positions other than the two shown, the mass has both kinetic and potential energy.)
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WORKED EXAMPLE 3.14
A body of mass 4.2 kg with initial speed 5.6 m sâ1 begins to move up an incline, as shown in Figure 3.25.

Figure 3.25: For worked example 3.14.

The body will be momentarily brought to rest after colliding with a spring of spring constant 2200 N mâ1. The body stops a vertical distance h = 0.85 m above its initial position. Determine the amount by which the spring has been compressed. Assume no frictional forces.
Answer
There are no external forces doing work, the system is isolated, and we have conservation of total mechanical energy.
Initially we have just kinetic energy, so
EintialÂ =12mv2+mgh+12kx2=12Ã4.2Ã5.62+0+0=65.856Â J
When the body stops we have gravitational and elastic potential energies
EtinalÂ =12mv2+mgh+12kx2 =0+4.2Ã9.81Ã0.85+12Ã2200Ãx2 =35.02+1100Â x2
Thus, equating Einitial to Efinal we find
35.02+1100Â x21100Â x2x2x=65.856=30.84=0.02804=0.17Â m
What happens when external forces do work on the system?
We have seen that in the absence of external forces doing work, the total mechanical energy, ET, of the system stays the same: we have the law of conservation of total mechanical energy. However, if there are interactions between the system and its surroundings (Figure 3.26) the total mechanical energy may change. These interactions mainly involve work done Wext by the surroundings and/or the transfer of thermal energy (heat) Q, to or from the surroundings.

Figure 3.26: The total energy of a system may change as a result of interactions with its surroundings.

So we expect that
ÎET = Wext + Q
In this chapter we will deal with Q = 0 so in that case we have the relation:
ÎET = Wext
EXAM TIP
You must make sure that you do not confuse the workâkinetic energy relation Wnet = âEK with âET = Wext. The workâkinetic energy relation relates the net work on a system to the change in the systemâs kinetic energy. The other relates the work done by external forces to the change of the total energy.
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WORKED EXAMPLE 3.15
A body of mass 2.0 kg (initially at rest) slides down a curved path of total length 22 m, as shown in Figure 3.27. The body starts from a vertical height of 5.0 m from the bottom. When it reaches the bottom, its speed is measured and found to equal 6.0 m sâ1.
aShow that there is a force resisting the motion.
bAssuming the force to have constant magnitude and opposite to the velocity, determine the magnitude of the force.

Figure 3.27: For worked example 3.15.

Answer
aWe calculate the total energy initially and finally to see if energy is conserved.
At the top:
EintialÂ =12mv2+mgh=0+2.0Ã9.8Ã5.0=98Â J
At the bottom:
EfinalÂ =12mv2+mgh=12Ã2.0Ã6.02+0=36Â J
The total energy has decreased, which shows the presence of an external frictional force resisting the motion.
bFrom âET = Wext we deduce that Wext = â62 J. This is the work done by the frictional force of magnitude f.
The force acts in the opposite direction to the velocity, so
âfsÃ(â1)=â62Â Jf=6222f=2.8Â N
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WORKED EXAMPLE 3.16
A mass of 5.00 kg moving up an incline with an initial velocity of 2.0 m sâ1 is acted upon by an external force of 55 N in the direction of the velocity. The motion is opposed by a frictional force. After travelling a distance of 18 m along the incline the mass has been raised to a vertical height of 3.0 m and the velocity of the body becomes 15 m sâ1. Determine the magnitude of the frictional force.
Answer
The change in the total mechanical energy ÎET is
ÎET=(12Ã5.0Ã152+5.0Ã9.81Ã3.0)â(12Ã5.0Ã2.02)=699.7Â J
Let the frictional force be f. The work done on the mass by the external forces is
(55âf)Ã18, and so from ÎET=WextÂ 
(55âf)Ã18=699.7
55âf=699.718
55âf=38.87
fâ16Â N
TEST YOUR UNDERSTANDING
22A ball is released from rest from the top of a box along three different ramps, X, Y and Z.

Figure 3.28: For question 22.

Along which ramp will the ball attain the greatest speed at the bottom?
23A ball is projected from a table twice, the first time horizontally and the second time at an angle of 45Â° to the horizontal. The speed is the same in both cases. In which case will the ball reach the ground with the greatest speed?
24Look at Figure 3.29.
aCalculate the minimum initial speed v the ball must have in order to make it to position B.
bUsing the answer in a calculate the speed at B.
cIf the initial speed is v = 12.0 m sâ1, calculate the speed at A and B.

Figure 3.29: For question 24.

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25Look at Figure 3.30. The speed of the 8.0 kg mass in position A is 6.0 m sâ1. By the time it gets to position B, its speed is 12.0 m sâ1.

Figure 3.30: For question 25.

Estimate the frictional force opposing the motion. (The frictional force is acting along the plane.)
26In a loop-the-loop toy, a small marble is released from rest from an initial position a height H above level ground. The radius of the loop is R. Frictional forces are negligible.

Figure 3.31: For question 26.

aShow that the speed of the marble at P is given by v=2g(Hâ2R)âââââââââââ.
bFind an expression for the normal force on the marble at P.
cDeduce (in terms of R) the minimum value of H for which the marble will not fall off the loop.
