Chapter 6.9 - Potential Energy of a Spring

In the previous section, we saw the law of conservation of energy. In this section, we will see the energy associated with a spring.

1. Consider a spring whose one end is attached to the ceiling. Fig.6.28(a) below:

Fig.6.28

• A red dashed horizontal line is drawn through the lower end of the spring
• This red dashed line is important for our calculations
2. In fig.b, a mass of 5 kg is attached to the lower end
• We can see that, now the lower end of the spring is 2 cm below the red line
3. The lower end of the spring coincided with the red line when no load was attached
• Now it is 2 cm below
■ So it is clear that, the spring is stretched by a displacement of 2 cm
• We denote this displacement as 'x'
4. We know that, for stretching or compressing a spring, force is required
• Here we have certainly applied a stretching force of (5 kg × g)
• Taking g as 10 ms^-2, we get: Stretching force F = 50 N
• We can write: When force F = 50 N, the displacement x of the spring = 2 cm
• Since we have to repeat the experiment, we will give special notations:
F₁ = 50 N, x₁ = 2 cm
5. Now we repeat the experiment
• We remove the 5 kg mass and attach a 12 kg mass. This is shown in fig.c
• We can see that, now the lower end of the spring is 4.8 cm below the red line
• The lower end of the spring coincided with the red line when no load was attached
• Now it is 4.8 cm below
• So it is clear that, the spring is stretched by a displacement (x) of 4.8 cm
• We can write: F₂ = 120 N, x₂ = 4.8 cm
6. We repeat the experiment one more time
• We remove the 12 kg mass and attach a 8 kg mass. This is shown in fig.d
• We can see that, now the lower end of the spring is 4.8 cm below the red line
• The lower end of the spring coincided with the red line when no load was attached
• Now it is 4.8 cm below
• So it is clear that, the spring is stretched by a displacement (x) of 4.8 cm
• We can write: F₃ = 120 N, x₃ = 3.2 cm
7. Now we find the ratio $\mathbf\small{\frac{F}{x}}$ for each case. We get:
• Experiment 1: = $\mathbf\small{\frac{F_1}{x_1}=\frac{50}{2}=25}$
• Experiment 2: = $\mathbf\small{\frac{F_2}{x_2}=\frac{120}{4.8}=25}$
• Experiment 3: = $\mathbf\small{\frac{F_3}{x_3}=\frac{80}{3.2}=25}$
8. We find that, the ratio is a constant for all combinations
• We can repeat the experiment for any number of times we like. We will get the same value of 25
• Since $\mathbf\small{\frac{F}{x}}$ is constant for a spring, it is give a special name: Spring constant
• It is denoted by k
• F in the numerator has unit N and x in the denominator has unit m. So the unit of 'k' is Nm^-1.
8. So we have $\mathbf\small{\frac{F}{x}=k}$
• If 'k' is very large, the left side $\mathbf\small{\frac{F}{x}}$ must also be very large
• For $\mathbf\small{\frac{F}{x}}$ to be large, the numerator F should be very large
    ♦ and denominator x should be very small
9. That means, if 'k' is very large, force F also will have to be very large even to produce a small x
■ Such a spring with large 'k' value is called a stiff spring
• If 'k' is small, F need not be so large to produce significant x
■ Such a spring with small 'k' value is called a soft spring.
■ Every spring has it's own unique value of 'k'   
10. Rearranging $\mathbf\small{\frac{F}{x}=k}$, we get: $\mathbf\small{F=kx}$
• Fom this expression, it is clear that, x is proportional to F

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• The above properties of spring were first discovered in 1660 by the British scientist Robert Hooke
• It is known by his name as: Hooke's Law
• In his book, he wrote: “as the extension, so the force”. That means, extension is proportional to the force. 
• This proportionality is clear from the equation F= kx. 
    ♦ When F increases, x also increases. 
    ♦ When F decreases, x also decreases
(Some basic lessons on proportionality can be seen here)
• However, Hookes' law is stated in the form: F = -kx

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• We will now see the reason for the negative sign:
We will write the steps:
1. In fig,6.29 (a) below, a block of mass m rests on a smooth horizontal plane

Fig.6.29

• It is attached to one end of a spring 
• The other end of the spring is rigidly attached to a vertical wall
2. In fig.a, the spring is in it’s normal configuration. That is., it is neither stretched nor compressed
• In this configuration, a red dashed vertical line is drawn through the 'point of contact' between the block and the spring
3. Since at this position, the spring is neither stretched or compressed, the displacement x is zero
• So we can write x = 0 along this dashed line
4. In fig.6.29(b), the block is pulled towards the right by a distance x
• So the spring is stretched by a distance x
5. To keep the spring in this stretched position, a force need to be applied
• We know that, this force is given by F = kx
• If we know the value of the spring constant k, we can multiply that k by the displacement x to obtain F
• So magnitude of F required to keep the block at the position in fig.b can be easily obtained
6. Now we will see the direction of F:
• When the spring is stretched by an external force F, an internal force (spring force F_s) of the same magnitude will develop inside the spring
• This F_s will try to restore it’s normal configuration
• That means, F_s will be opposite in direction to F
7. We applied F towards the right. That is., towards the positive side of the x axis
• So F can be considered to be positive
■ Then internal force F_s will be negative
8. Now we will consider compression. In fig.c, the block is pushed towards the left by a distance x
• So the spring is compressed by a distance x
9. To keep the spring in this compressed position, a force need to be applied
• We know that, this force is given by F = kx
• If we know the value of the spring constant k, we can multiply it with the displacement x to obtain F
• So magnitude of F required to keep the block at the position in fig.c can be easily obtained
10. Now we will see the direction of F:
• When the spring is compressed by an external force F, an internal force (spring force F_s) of the same magnitude will develop inside the spring
• This F_s will try to restore it’s normal configuration
• That means, F_s will be opposite in direction to F
11. We applied F towards the left. That is., towards the negative side of the x axis
• So F can be considered to be negative
■ Then internal force F_s will be positive
12. So we have a situation in which two conditions are involved:
(i) From (7) we have:
When x is positive, F_s is negative
(ii) From (11) we have:
When x is negative, F_s is positive
■ To satisfy both the conditions, we must write: F_s = -kx

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Now we know the reason for the negative sign. Next we will see the peculiarities of this relation

We will write the steps:
1. In the relation F_s = -kx:
• The left side gives the internal force developed within the spring
• We enter the displacement x on the right side
2. When x changes, F_s will also change
• So this relation can be used to plot the Force-Displacement graph
• What will be the shape of this graph?
3. In the case of gravitational force, the force does not change with 'distance from the ground'
• Remember that, we plot displacement along x axis and force along y axis
• So the force-displacement graph in the case of gravity will be a horizontal line 
4. But the spring force varies with distance
• Luckily for us, it is a uniform variation. Because, force is proportional to displacement
• For uniform variation, the graph will be a straight line
    ♦ But this straight line will be neither horizontal nor vertical
    ♦ It will be inclined
(If it is not a uniform variation, the graph will be a curve)
5.So our present graph showing the relation F_s = -kx will be an inclined straight line
• Let us see it’s peculiarities:
(i) From analytic geometry classes, we know that the equation of a straight line is y = mx +c
• If the constant c = 0, then y intercept will be zero
• That means, the straight line will then pass through the origin
• Further, if the slope m is negative, the line will slope downwards (with increase in x)
(ii) So F_s = -kx is the equation of a line which pass through the origin
• Also, with the increase in x, it will be sloping downwards
• This is shown by the cyan line in fig.d

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Next we will find the work done by F_s when the spring is stretched :

1. We know that work done by a force is equal to the ‘area enclosed by the force-displacement graph. (Details here)
2. Let the spring be stretched by a distance x_t as shown in the fig.6.30(a) below

Fig.6.30

• Then the work done by F_s will be equal to the area of the red triangle shown in the fig.6.30(b)
3. It is easy to calculate this area. We have:
• Base OP of the red triangle = x_t
• Altitude PQ of the red triangle = Magnitude of F_s at x_t
    ♦ The relation F_s = kx gives the magnitude of Fs at any x
    ♦ So F_s at x_t = k×x_t
■ So area = ¹⁄₂ × base × altitude = $\mathbf\small{\frac{1}{2}\times x_t \times k \times x_t=\frac{k\,x_t^2}{2}}$
■ Thus we can write:
Work done by the spring force F_s when the spring is stretched through a distance of x_t = $\mathbf\small{\frac{k\,x_t^2}{2}}$
4. When we discussed about gravitational potential energy (P_g), we saw the following 7 points:
(i) Gravitational force is the original force. With out that, there is no P_g
(ii) So work done by gravity is taken as positive
(iii) When an object is raised to a height by an external force, work is done against gravity. 
(iv) So work done by the external force is taken as negative
(v) That means., the P_g stored in an object (of mass m) at a height h = -mgh joules
(vi) When the object falls, work is done by gravity
(vii) So the amount of energy released when the object is in free fall is +mgh joules
5. We can follow the same sign convention here also
• F_s and the external force F are numerically equal. So work done by F_s is numerically equal to the work done by external force F
• Since F is opposite to the original force F_s, we can write:
■ Work done by F when the spring is stretched (from normal configuration) by x_t = $\mathbf\small{-\frac{k\,x_t^2}{2}}$
■ Work done by F_s when the spring is stretched (from normal configuration) by x_t = $\mathbf\small{\frac{k\,x_t^2}{2}}$
• This much work is stored in the spring as ‘spring potential energy’

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Next we will find the work done by F_s when the spring is compressed:

1. We know that work done by a force is equal to the ‘area enclosed by the force-displacement graph
2. Let the spring be compressed by a distance x_c as shown in the fig.6.31(a) below

Fig.6.31

• Then the work done by F_s will be equal to the area of the magenta triangle shown in the fig.6.31(b)
3. It is easy to calculate this area. We have:
• Base OR of the magenta triangle = x_c
• Altitude RS of the magenta triangle = Magnitude of F_s at x_c
    ♦ The relation F_s = kx gives the magnitude of Fs at any x
    ♦ So F_s at x_c = k×x_c
■ So area = ¹⁄₂ × base × altitude = $\mathbf\small{\frac{1}{2}\times x_c \times k \times x_c=\frac{k\,x_c^2}{2}}$
■ Thus we can write:
Work done by the spring force F_s when the spring is compressed through a distance of x_c = $\mathbf\small{\frac{k\,x_c^2}{2}}$
4. We can follow the same sign convention as the gravitational potential energy, here also
• F_s and the external force F are numerically equal. So work done by F_s is numerically equal to the work done by external force F
• Since F is opposite to the original force F_s, we can write:
■ Work done by F when the spring is compressed (from normal configuration) by x_c = $\mathbf\small{-\frac{k\,x_c^2}{2}}$
■ Work done by F_s when the spring is compressed (from normal configuration) by x_c = $\mathbf\small{\frac{k\,x_c^2}{2}}$
• This much work is stored in the spring as ‘spring potential energy’

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In the next section, we will prove that spring force is a conservative force. And hence, the spring energy follows the law of conservation of energy.

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Chapter 6.8 - The Conservation of Mechanical Energy

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:

Fig.6.25

• 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

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4. When the block is at A:

    ♦ 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

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7. When the block is at A:

    ♦ 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

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10. From the results in (3), (6) and (9), we can write:

■ 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

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The law of conservation of energy states that:

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

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• 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

• 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 = K_f - K_i = 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 (P_f) after the displacement of Δx 
= Initial potential energy + the change in potential energy
= P_i + ΔP 
■ Total kinetic energy (K_f) after the displacement of Δx
= Initial kinetic energy + the change in kinetic energy
= K_i + ΔK
7. So total energy at that point (P_f + K_f) = [(P_i + ΔP) + (K_i + ΔK)] = [(P_i + K_i) + (ΔP + ΔK)]
• Since ΔP and ΔK are numerically equal but opposite in signs, they will cancel each other
• So we get: P_f + K_f = P_i + K_i
■ 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

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Now we will see some solved examples:

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]

Fig.6.26

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:

Fig.6.27

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., P_i + K_i = P_f + K_f
$\mathbf\small{\Longrightarrow [mg(h+32)+0]=[mg(h+7)+\frac{1}{2}mv_c^2]}$
• 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

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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:

4. P_i + K_i = P_f + K_f
$\mathbf\small{\Longrightarrow [mg \times 32+0]=[mg \times 7+\frac{1}{2}mv_c^2]}$
• Where 'm' is the mass of the trolley. But it will cancel out from either sides of the equation
• So we get: $\mathbf\small{[32g]=[7g+0.5v_c^2]}$
$\mathbf\small{\Longrightarrow 25 \times 9.8=0.5v_c^2}$
$\mathbf\small{\Longrightarrow v_c=22.136\;ms^{-1}}$

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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:
P_i + K_i is equal to P_f + K_f
• But the unknown resistive force does some 'negative work (W_r)' on the drop. 
    ♦ This W_r 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 (W_r) to the final energy, a new equality can be established
• That is: P_i + K_i = P_f + K_f + W_r.
• 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: P_i + K_i = P_f + K_f + W_r
$\mathbf\small{\Longrightarrow [mg \times 1000+0]=[0+\frac{1}{2}mv^2]+W_r}$
$\mathbf\small{\Longrightarrow [0.001 \times 10 \times 1000+0]=[0+0.5 \times 0.001 \times 50^2]+W_r}$
$\mathbf\small{\Longrightarrow [10]=[1.25]+W_r}$
• Thus we get: W_r = (10-1.25) = 8.75 joules
• This is the same result that we obtained before

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In the next section, we will see how energy can be stored in springs

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Chapter 6.7 - Potential energy can change into Kinetic energy

In the previous section, we saw potential energy. In this section, we will see the relation between potential energy and kinetic energy.

1. In fig.6.25 below, a block of mass 'm' is at point A which is at a height of h from the ground

Fig.6.25

• So the block has a potential energy of -mgh joules
• That is., P_g(A) = -mgh 
2. Let the block be allowed to fall freely from A
• Consider the instant when it is at B
• At that instant, it's potential energy will be -mgh_{1
• That is., Pg(B) =} -mgh₁
3. -mgh₁ is lesser (in magnitude) than mgh because, h₁ is less than h
• So it is clear that, some potential energy is lost. 
• The quantity of 'potential energy lost' = P_g(B) - P_g(A) = -mgh₁-(-mgh) = -mgh₁ + mgh = mg(h-h₁)
4. So there is a loss of mg(h-h₁) joules
• But where did mg(h-h₁) joules go?
Let us analyze:
5. Let us find the velocity at the instant when the block is at B:
• We can use the familiar equation: $\mathbf\small{v^2-v_0^2=2ax}$
• Substituting known values, we get: $\mathbf\small{v_B^2=2g(h-h_1)}$
6. So we get 'square of the velocity at B'
• No need to find the actual velocity by taking square root. Because, square is what we want
• Multiplying this square by $\mathbf\small{\frac{1}{2}m}$, we get:
$\mathbf\small{\frac{1}{2}m \times v_B^2=\frac{1}{2}m \times [2g(h-h_1)]=mg(h-h_1)}$
7. Look at the first and last terms of the above result
• The first term is the kinetic energy at B
• The last term is the Pg lost due to the lowering from A to B
8. So we can write:
The 'P_g lost by the block due to lowering from A to B' is equal to the 'kinetic energy acquired at B'
• That means, the 'P_g lost' did not go any where. It just changed to another form of energy

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Let us confirm this by checking at another point C
We will write all the steps again:
1. In fig.6.25 above, a block of mass 'm' is at point A which is at a height of h from the ground. 
• So the block has a potential energy of -mgh joules
• That is., P_g(A) = -mgh 
2. Let the block be allowed to fall freely from A
• Consider the instant when it is at C
• At that instant, it's potential energy will be -mgh_{2
• That is., Pg(C) =} -mgh₂
3. -mgh₂ is lesser (in magnitude) than mgh because, h₂ is less than h
• So it is clear that, some potential energy is lost. 
• The quantity of 'potential energy lost' = P_g(C) - P_g(A) = -mgh₂-(-mgh) = -mgh₂ + mgh = mg(h-h₂)
4. So there is a loss of mg(h-h₂) joules
• But where did mg(h-h₂) joules go?
Let us analyze:
5. Let us find the velocity at the instant when the block is at C:
• We can use the familiar equation: $\mathbf\small{v^2-v_0^2=2ax}$
• Substituting known values, we get: $\mathbf\small{v_C^2=2g(h-h_2)}$
6. So we get 'square of the velocity at C'
• No need to find the actual velocity by taking square root. Because, square is what we want
• Multiplying this square by $\mathbf\small{\frac{1}{2}m}$, we get:
$\mathbf\small{\frac{1}{2}m \times v_C^2=\frac{1}{2}m \times [2g(h-h_2)]=mg(h-h_2)}$
7. Look at the first and last terms of the above result
• The first term is the kinetic energy at C
• The last term is the Pg lost due to the lowering from A to C
8. So we can write:
The 'P_g lost by the block due to lowering from A to C' is equal to the 'kinetic energy acquired at C'

• That means, the 'P_g lost' did not go any where. It just changed to another form of energy

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So it is confirmed. Let us see what happens at the instant when the block reaches the ground point D
We will write all the steps again:
1. In fig.6.25 above, a block of mass 'm' is at point A which is at a height of h from the ground. 
• So the block has a potential energy of -mgh joules
• That is., P_g(A) = -mgh 
2. Let the block be allowed to fall freely from A
• Consider the instant when it is at D
• At that instant, it's potential energy will be 0
_{• That is., Pg(D) = 0}
3. 0 is lesser (in magnitude) than mgh
• So it is clear that, some potential energy is lost. 
• The quantity of 'potential energy lost' = P_g(D) - P_g(A) = 0-(-mgh) = mgh
4. So there is a loss of mgh joules
• That means, all the potential energy mgh, which was available at A is lost
• But where did mgh joules go?
Let us analyze:
5. Let us find the velocity at the instant when the block is at D:
• We can use the familiar equation: $\mathbf\small{v^2-v_0^2=2ax}$
• Substituting known values, we get: $\mathbf\small{v_D^2=2gh}$
6. So we get 'square of the velocity at D'
• No need to find the actual velocity by taking square root. Because, square is what we want
• Multiplying this square by $\mathbf\small{\frac{1}{2}m}$, we get:
$\mathbf\small{\frac{1}{2}m \times v_D^2=\frac{1}{2}m \times [2gh]=mgh}$
7. Look at the first and last terms of the above result
• The first term is the kinetic energy at D
• The last term is the Pg lost due to the lowering from A to D
8. So we can write:
The 'P_g lost by the block due to lowering from A to D' is equal to the 'kinetic energy acquired at D'
• That means, the 'P_g lost' did not go any where. It just changed to another form of energy
9. Thus, at the lower most point, all the potential energy is converted into kinetic energy

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We have seen work-energy theorem in a previous section. (Details here)
Let us apply it to our present case:
■ Consider the displacement from A to B
We will write the steps:
1. Initial kinetic energy = kinetic energy at A = K_i = 0
2. Final kinetic energy = kinetic energy at B = K_f = mg(h-h₁) [From the first step (6) above]
3. So 'change in kinetic energy = K_f-K_i = mg(h-h₁)
4. Work done by gravity from A to B = mg × distance AB = mg(h-h₁)
5. The results in (3) and (4) are the same. So displacement from A to B satisfies the work-energy theorem

■ Consider the displacement from B to C
We will write the steps:
1. Initial kinetic energy = kinetic energy at B = K_i = mg(h-h₁) [From the first step (6) above]
2. Final kinetic energy = kinetic energy at C = K_f = mg(h-h₂) [From the second step (6) above]
3. So 'change in kinetic energy = K_f-K_i = [mg(h-h₂)-mg(h-h₁)] = [mgh-mgh₂-mgh+mgh₁] = mg(h₁-h₂)
4. Work done by gravity from B to C = mg × distance BC = mg(h₁-h₂)
5. The results in (3) and (4) are the same. So displacement from B to C satisfies the work-energy theorem

■ Consider the displacement from C to D
We will write the steps:
1. Initial kinetic energy = kinetic energy at C = K_i = mg(h-h₂) [From the second step (6) above]
2. Final kinetic energy = kinetic energy at D = K_f = mgh [From the third step (6) above]
3. So 'change in kinetic energy = K_f-K_i = [mgh-mg(h-h₂)] = [mgh-mgh+mgh₂] = mgh₂
4. Work done by gravity from C to D = mg × distance CD = mgh₂
5. The results in (3) and (4) are the same. So displacement from C to D satisfies the work-energy theorem

■ Consider the total displacement from A to D
We will write the steps:
1. Initial kinetic energy = kinetic energy at A = K_i = 0
2. Final kinetic energy = kinetic energy at D = K_f = mgh [From the third step (6) above]
3. So 'change in kinetic energy = K_f-K_i = [mgh-0] = mgh
4. Work done by gravity from A to D = mg × distance AD = mgh
5. The results in (3) and (4) are the same. So displacement from A to D satisfies the work-energy theorem

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So we find that a freely falling body satisfies work-energy theorem at whichever segment that we consider along it's path. In the next section, we will see conservation of energy

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