In the previous section, we saw the change of potential energy into kinetic energy. In this section, we will see the conservation of energy.
We will write the steps:
1. Consider the fig.6.25 that we saw in the previous section. For convenience, it is shown again below:
• When the block is at A:
♦ Potential energy = mgh
♦ Kinetic energy = 0
So total energy at the initial position A = mgh + 0 = mgh joules
• When the block is at B:
♦ Potential energy becomes a value lesser than mgh
♦ That means, there is a loss in potential energy
♦ Also, kinetic energy increases to a value above zero
♦ That means, there is a gain in kinetic energy
2. We saw that the 'loss in potential energy' is equal to the 'gain in kinetic energy'
Let us write the exact values:
• Exact potential energy at B = mgh1
• Exact kinetic energy at B = $\mathbf\small{\frac{1}{2}m \times v_B^2=\frac{1}{2}m \times [2g(h-h_1)]=mg(h-h_1)}$ [From previous section]
3. So total energy at B = mgh1 +[mg(h-h1)] = mgh
♦ Potential energy = mgh
♦ Kinetic energy = 0
• When the block is at C:
♦ Potential energy becomes a value lesser than mgh
♦ That means, there is a loss in potential energy
♦ Also, kinetic energy increases to a value above zero
♦ That means, there is a gain in kinetic energy
5. We saw that the 'loss in potential energy' is equal to the 'gain in kinetic energy'
Let us write the exact values:
• Exact potential energy at C = mgh2
• Exact kinetic energy at C = $\mathbf\small{\frac{1}{2}m \times v_C^2=\frac{1}{2}m \times [2g(h-h_2)]=mg(h-h_2)}$ [From previous section]
6. So total energy at C = mgh2 +[mg(h-h2)] = mgh
♦ Potential energy = mgh
♦ Kinetic energy = 0
• When the block is at D:
♦ Potential energy becomes a value lesser than mgh
♦ That means, there is a loss in potential energy
♦ Also, kinetic energy increases to a value above zero
♦ That means, there is a gain in kinetic energy
8. We saw that the 'loss in potential energy' is equal to the 'gain in kinetic energy'
Let us write the exact values:
• Exact potential energy at D = 0
• Exact kinetic energy at C = $\mathbf\small{\frac{1}{2}m \times v_D^2=\frac{1}{2}m \times [2g(h-0)]=mgh}$ [From previous section]
9. So total energy at D = 0 +[mgh] = mgh
■ Total energy of the block remains unchanged
• The block started off from the point A with a total energy of mgh joules
• We can take any point along it's path. The total energy at that point will be the same mgh joules
Energy can neither be created nor destroyed. It transforms from one form to another form. However the total energy remains constant
• This law will work only when all the forces acting are conservative forces.
• If non-conservative forces like friction or air resistance also act, some energy will be lost as heat, sound etc.,
• Then the total energy will not remain constant
• Let us write a general form so that, the law can be applied to all conservative forces
• We will write the steps:
1. Consider a conservative force F acting on a body
• Let the body be displaced through Δx
• So work done by F = F×Δx
2. Using work-energy theorem, we have:
• Change in kinetic energy of the body = work done
• So that: ΔK = Kf - Ki = F×Δx
3. Thus we get an expression for the 'change in kinetic energy'
• Now we want an expression for the 'change in potential energy'
4. We have seen that ‘quantity of potential energy lost’ is equal to ‘quantity of kinetic energy gained’
• ‘Quantity of kinetic energy gained’ is the ‘change in kinetic energy’
• But as mentioned in (2), the ‘change in kinetic energy’ is equal to the ‘work done’ F×Δx
■ So we get:
quantity of potential energy lost (ΔP)
= quantity of kinetic energy gained (ΔK)
= work done (F×Δx)
• So both ΔP and ΔK are equal to F×Δx
5. But ΔP and ΔK are equal only in magnitudes. The signs are different
• Let us see the reason:
♦ P decreases from a higher value to a lower value
♦ K increases from a lower value to higher value
• So we can write: ΔP = - ΔK
6. Now we write the total energies:
■ Total potential energy (Pf) after the displacement of Δx
= Initial potential energy + the change in potential energy
= Pi + ΔP
■ Total kinetic energy (Kf) after the displacement of Δx
= Initial kinetic energy + the change in kinetic energy
= Ki + ΔK
7. So total energy at that point (Pf + Kf) = [(Pi + ΔP) + (Ki + ΔK)] = [(Pi + Ki) + (ΔP + ΔK)]
• Since ΔP and ΔK are numerically equal but opposite in signs, they will cancel each other
• So we get: Pf + Kf = Pi + Ki
■ That means 'total final energy' is same as 'total initial energy'
■ Thus we get a general form for the law of conservation of energy
■ The sum of potential energy and kinetic energy is called the total mechanical energy of the system
Solved example 6.15
The trolley in fig.6.26 below, is initially at rest at the top most point A of the hill. It is allowed to travel down from point A. What is the velocity of the trolley when it passes point C? Neglect the effect of friction. [Take g = 9.8 ms-2]
Solution:
1. Point A is 32 m above B. Point C is 7 m above B
• We are not given the height of B from the surface of the earth
• Let us assume that B is ‘h’ m above the surface of the earth. This is shown in fig.6.27 below:
2. When the trolley reaches B, it would have lost a lot of potential energy
• But all that ‘lost potential energy’ will be converted into kinetic energy
• That means, the trolley will have a large velocity when it reaches B
• This kinetic energy will enable the trolley to climb upwards on the other side and reach C
3. We need to consider the initial and final points (A and C) only. This is because, work done is independent of the path. It depends on the initial and final points only
• We can apply the law of conservation of energy:
Total mechanical energy at A = Total mechanical energy at C
4. That is., Pi + Ki = Pf + Kf
$\mathbf\small{\Longrightarrow [mg(h+32)+0]=[mg(h+7)+\frac{1}{2}mv_c^2]}$
• Where 'm' is the mass of the trolley. But it will cancel out from either sides of the equation
• So we get: $\mathbf\small{[g(h+32)]=[g(h+7)+0.5v_c^2]}$
$\mathbf\small{\Longrightarrow [gh+32g]=[gh+7g+0.5v_c^2]}$
$\mathbf\small{\Longrightarrow 25 \times 9.8=0.5v_c^2}$
$\mathbf\small{\Longrightarrow v_c=22.136\;ms^{-1}}$
• Note that, the unknown 'h' also cancels out in an intermediate step
4. Pi + Ki = Pf + Kf
$\mathbf\small{\Longrightarrow [mg \times 32+0]=[mg \times 7+\frac{1}{2}mv_c^2]}$
• Where 'm' is the mass of the trolley. But it will cancel out from either sides of the equation
• So we get: $\mathbf\small{[32g]=[7g+0.5v_c^2]}$
$\mathbf\small{\Longrightarrow 25 \times 9.8=0.5v_c^2}$
$\mathbf\small{\Longrightarrow v_c=22.136\;ms^{-1}}$
We will write the steps:
1. Consider the fig.6.25 that we saw in the previous section. For convenience, it is shown again below:
 |
| Fig.6.25 |
♦ Potential energy = mgh
♦ Kinetic energy = 0
So total energy at the initial position A = mgh + 0 = mgh joules
• When the block is at B:
♦ Potential energy becomes a value lesser than mgh
♦ That means, there is a loss in potential energy
♦ Also, kinetic energy increases to a value above zero
♦ That means, there is a gain in kinetic energy
2. We saw that the 'loss in potential energy' is equal to the 'gain in kinetic energy'
Let us write the exact values:
• Exact potential energy at B = mgh1
• Exact kinetic energy at B = $\mathbf\small{\frac{1}{2}m \times v_B^2=\frac{1}{2}m \times [2g(h-h_1)]=mg(h-h_1)}$ [From previous section]
3. So total energy at B = mgh1 +[mg(h-h1)] = mgh
4. When the block is at A:
♦ Kinetic energy = 0
• When the block is at C:
♦ Potential energy becomes a value lesser than mgh
♦ That means, there is a loss in potential energy
♦ Also, kinetic energy increases to a value above zero
♦ That means, there is a gain in kinetic energy
5. We saw that the 'loss in potential energy' is equal to the 'gain in kinetic energy'
Let us write the exact values:
• Exact potential energy at C = mgh2
• Exact kinetic energy at C = $\mathbf\small{\frac{1}{2}m \times v_C^2=\frac{1}{2}m \times [2g(h-h_2)]=mg(h-h_2)}$ [From previous section]
6. So total energy at C = mgh2 +[mg(h-h2)] = mgh
7. When the block is at A:
♦ Kinetic energy = 0
• When the block is at D:
♦ Potential energy becomes a value lesser than mgh
♦ That means, there is a loss in potential energy
♦ Also, kinetic energy increases to a value above zero
♦ That means, there is a gain in kinetic energy
8. We saw that the 'loss in potential energy' is equal to the 'gain in kinetic energy'
Let us write the exact values:
• Exact potential energy at D = 0
• Exact kinetic energy at C = $\mathbf\small{\frac{1}{2}m \times v_D^2=\frac{1}{2}m \times [2g(h-0)]=mgh}$ [From previous section]
9. So total energy at D = 0 +[mgh] = mgh
10. From the results in (3), (6) and (9), we can write:
• The block started off from the point A with a total energy of mgh joules
• We can take any point along it's path. The total energy at that point will be the same mgh joules
The law of conservation of energy states that:
• This law will work only when all the forces acting are conservative forces.
• If non-conservative forces like friction or air resistance also act, some energy will be lost as heat, sound etc.,
• Then the total energy will not remain constant
• In the above discussion, we learnt about the law of conservation of energy by taking the 'force of gravity' as an example. The 'force of gravity' is a conservative force
• We will write the steps:
1. Consider a conservative force F acting on a body
• Let the body be displaced through Δx
• So work done by F = F×Δx
2. Using work-energy theorem, we have:
• Change in kinetic energy of the body = work done
• So that: ΔK = Kf - Ki = F×Δx
3. Thus we get an expression for the 'change in kinetic energy'
• Now we want an expression for the 'change in potential energy'
4. We have seen that ‘quantity of potential energy lost’ is equal to ‘quantity of kinetic energy gained’
• ‘Quantity of kinetic energy gained’ is the ‘change in kinetic energy’
• But as mentioned in (2), the ‘change in kinetic energy’ is equal to the ‘work done’ F×Δx
■ So we get:
quantity of potential energy lost (ΔP)
= quantity of kinetic energy gained (ΔK)
= work done (F×Δx)
• So both ΔP and ΔK are equal to F×Δx
5. But ΔP and ΔK are equal only in magnitudes. The signs are different
• Let us see the reason:
♦ P decreases from a higher value to a lower value
♦ K increases from a lower value to higher value
• So we can write: ΔP = - ΔK
6. Now we write the total energies:
■ Total potential energy (Pf) after the displacement of Δx
= Initial potential energy + the change in potential energy
= Pi + ΔP
■ Total kinetic energy (Kf) after the displacement of Δx
= Initial kinetic energy + the change in kinetic energy
= Ki + ΔK
7. So total energy at that point (Pf + Kf) = [(Pi + ΔP) + (Ki + ΔK)] = [(Pi + Ki) + (ΔP + ΔK)]
• Since ΔP and ΔK are numerically equal but opposite in signs, they will cancel each other
• So we get: Pf + Kf = Pi + Ki
■ That means 'total final energy' is same as 'total initial energy'
■ Thus we get a general form for the law of conservation of energy
■ The sum of potential energy and kinetic energy is called the total mechanical energy of the system
Now we will see some solved examples:
The trolley in fig.6.26 below, is initially at rest at the top most point A of the hill. It is allowed to travel down from point A. What is the velocity of the trolley when it passes point C? Neglect the effect of friction. [Take g = 9.8 ms-2]
 |
| Fig.6.26 |
1. Point A is 32 m above B. Point C is 7 m above B
• We are not given the height of B from the surface of the earth
• Let us assume that B is ‘h’ m above the surface of the earth. This is shown in fig.6.27 below:
 |
| Fig.6.27 |
• But all that ‘lost potential energy’ will be converted into kinetic energy
• That means, the trolley will have a large velocity when it reaches B
• This kinetic energy will enable the trolley to climb upwards on the other side and reach C
3. We need to consider the initial and final points (A and C) only. This is because, work done is independent of the path. It depends on the initial and final points only
• We can apply the law of conservation of energy:
Total mechanical energy at A = Total mechanical energy at C
4. That is., Pi + Ki = Pf + Kf
$\mathbf\small{\Longrightarrow [mg(h+32)+0]=[mg(h+7)+\frac{1}{2}mv_c^2]}$
• Where 'm' is the mass of the trolley. But it will cancel out from either sides of the equation
• So we get: $\mathbf\small{[g(h+32)]=[g(h+7)+0.5v_c^2]}$
$\mathbf\small{\Longrightarrow [gh+32g]=[gh+7g+0.5v_c^2]}$
$\mathbf\small{\Longrightarrow 25 \times 9.8=0.5v_c^2}$
$\mathbf\small{\Longrightarrow v_c=22.136\;ms^{-1}}$
• Note that, the unknown 'h' also cancels out in an intermediate step
We can solve the problem by considering point B to be the datum. That is., we assume that, the point B is on the earth surface. We need to redo step (4) only:
$\mathbf\small{\Longrightarrow [mg \times 32+0]=[mg \times 7+\frac{1}{2}mv_c^2]}$
• Where 'm' is the mass of the trolley. But it will cancel out from either sides of the equation
• So we get: $\mathbf\small{[32g]=[7g+0.5v_c^2]}$
$\mathbf\small{\Longrightarrow 25 \times 9.8=0.5v_c^2}$
$\mathbf\small{\Longrightarrow v_c=22.136\;ms^{-1}}$
Now let us redo solved example 6.4 which we did by applying principles of kinetic energy alone. This time we will apply the law of conservation of energy
Solved example 6.4 (Alternate method using consrvation of energy)
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height of 1 km. It hits the ground with a speed of 50.0 m s-1. (a) What is the work done by the gravitational force? (b) What is the work done by the unknown resistive force?
Solution:
• Originally we have an equality:
Pi + Ki is equal to Pf + Kf
• But the unknown resistive force does some 'negative work (Wr)' on the drop.
♦ This Wr will take away some energy from the drop.
♦ As a result, the 'final total energy' will not be equal to the 'initial total energy'
• If we add this unknown work (Wr) to the final energy, a new equality can be established
• That is: Pi + Ki = Pf + Kf + Wr.
• Now we can write the steps:
1. The initial position of the rain drop is 1000 m above the ground
• The final position is a point on the ground
2. We have: Pi + Ki = Pf + Kf + Wr
$\mathbf\small{\Longrightarrow [mg \times 1000+0]=[0+\frac{1}{2}mv^2]+W_r}$
$\mathbf\small{\Longrightarrow [0.001 \times 10 \times 1000+0]=[0+0.5 \times 0.001 \times 50^2]+W_r}$
$\mathbf\small{\Longrightarrow [10]=[1.25]+W_r}$
• Thus we get: Wr = (10-1.25) = 8.75 joules
• This is the same result that we obtained before
Solved example 6.4 (Alternate method using consrvation of energy)
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height of 1 km. It hits the ground with a speed of 50.0 m s-1. (a) What is the work done by the gravitational force? (b) What is the work done by the unknown resistive force?
Solution:
• Originally we have an equality:
Pi + Ki is equal to Pf + Kf
• But the unknown resistive force does some 'negative work (Wr)' on the drop.
♦ This Wr will take away some energy from the drop.
♦ As a result, the 'final total energy' will not be equal to the 'initial total energy'
• If we add this unknown work (Wr) to the final energy, a new equality can be established
• That is: Pi + Ki = Pf + Kf + Wr.
• Now we can write the steps:
1. The initial position of the rain drop is 1000 m above the ground
• The final position is a point on the ground
2. We have: Pi + Ki = Pf + Kf + Wr
$\mathbf\small{\Longrightarrow [mg \times 1000+0]=[0+\frac{1}{2}mv^2]+W_r}$
$\mathbf\small{\Longrightarrow [0.001 \times 10 \times 1000+0]=[0+0.5 \times 0.001 \times 50^2]+W_r}$
$\mathbf\small{\Longrightarrow [10]=[1.25]+W_r}$
• Thus we get: Wr = (10-1.25) = 8.75 joules
• This is the same result that we obtained before
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