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
• So the block has a potential energy of -mgh joules
• That is., Pg(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 -mgh1
• 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' = Pg(B) - Pg(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 'Pg lost by the block due to lowering from A to B' is equal to the 'kinetic energy acquired at B'
• That means, the 'Pg lost' did not go any where. It just changed to another form of energy
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., Pg(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 -mgh2
• 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' = Pg(C) - Pg(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 'Pg lost by the block due to lowering from A to C' is equal to the 'kinetic energy acquired at C'
• That means, the 'Pg lost' did not go any where. It just changed to another form of energy
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., Pg(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' = Pg(D) - Pg(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 'Pg lost by the block due to lowering from A to D' is equal to the 'kinetic energy acquired at D'
• That means, the 'Pg 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
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 = Ki = 0
2. Final kinetic energy = kinetic energy at B = Kf = mg(h-h1) [From the first step (6) above]
3. So 'change in kinetic energy = Kf-Ki = 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 = Ki = mg(h-h1) [From the first step (6) above]
2. Final kinetic energy = kinetic energy at C = Kf = mg(h-h2) [From the second step (6) above]
3. So 'change in kinetic energy = Kf-Ki = [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 = Ki = mg(h-h2) [From the second step (6) above]
2. Final kinetic energy = kinetic energy at D = Kf = mgh [From the third step (6) above]
3. So 'change in kinetic energy = Kf-Ki = [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 = Ki = 0
2. Final kinetic energy = kinetic energy at D = Kf = mgh [From the third step (6) above]
3. So 'change in kinetic energy = Kf-Ki = [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
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 |
• That is., Pg(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 -mgh1
• 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' = Pg(B) - Pg(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 'Pg lost by the block due to lowering from A to B' is equal to the 'kinetic energy acquired at B'
• That means, the 'Pg lost' did not go any where. It just changed to another form of energy
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., Pg(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 -mgh2
• 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' = Pg(C) - Pg(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 'Pg lost by the block due to lowering from A to C' is equal to the 'kinetic energy acquired at C'
• That means, the 'Pg lost' did not go any where. It just changed to another form of energy
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., Pg(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' = Pg(D) - Pg(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 'Pg lost by the block due to lowering from A to D' is equal to the 'kinetic energy acquired at D'
• That means, the 'Pg 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
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 = Ki = 0
2. Final kinetic energy = kinetic energy at B = Kf = mg(h-h1) [From the first step (6) above]
3. So 'change in kinetic energy = Kf-Ki = 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 = Ki = mg(h-h1) [From the first step (6) above]
2. Final kinetic energy = kinetic energy at C = Kf = mg(h-h2) [From the second step (6) above]
3. So 'change in kinetic energy = Kf-Ki = [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 = Ki = mg(h-h2) [From the second step (6) above]
2. Final kinetic energy = kinetic energy at D = Kf = mgh [From the third step (6) above]
3. So 'change in kinetic energy = Kf-Ki = [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 = Ki = 0
2. Final kinetic energy = kinetic energy at D = Kf = mgh [From the third step (6) above]
3. So 'change in kinetic energy = Kf-Ki = [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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