In the previous section, we saw the quantity of energy that will be stored in a spring when it is stretched or compressed. In this section, we will prove that spring energy is a conservative energy.
1. Consider a spring stretched by xt as shown in fig.6.32(a) below
• We know how to calculate the work required for stretching by xt:
Work required = Area of the red triangle
2. In some cases, we may want to know 'intermediate works'. This can be explained as follows:
• In fig.a, two points xi and xf are marked on the x axis
• We want to know how much work is required for stretching the spring from xi to xf
3. Obviously, this work is equal to the area of the shaded trapezium shown in fig.b
• The area of that trapezium can be calculated easily. All we need are the following 3 items:
(i) Height of the trapezium at xi
♦ This is equal to k xi
(ii) Height of the trapezium at xf
♦ This is equal to k xf
(iii) Base of the trapezium = (xf-xi)
• With the above 3 items, we will get the area. (Details about area of trapezium can be seen here)
■ So the work done will depend on the initial and final positions (xi and xf) only
4. What if, the path contains two segments as follows:
(i) The mass is moved from xi to xf
(ii) The mass is moved from xf to xi
• This the path is a cyclic path. The work done in segment-1 will be released in segment-2
■ Then the net work will be zero
5. So spring force is indeed a conservative force
• Since it is a conservative force, the energy stored in it (the spring potential energy) will follow the law of conservation of energy.
• So we can get back 'all the work done' at a later time
Now we will put this information into a practical application:
1. In fig.6.33(a) below, a block is pulled to a point A. It is kept at that position
• The distance of A from the equilibrium position 'O' is 'xt'
• So a potential energy of $\mathbf\small{\frac{k\,x_t^2}{2}}$ will be stored in the block
2. At that position, the block is stationary. So it's kinetic energy K is zero
• If we release the block, the spring will pull it towards the equilibrium position 'O'
• We can say this:
When released, the block moves towards 'O'
3. During this motion, the block is acted upon by the force of the spring
(This is just like a freely falling body acted upon by gravitational force)
• Since the block is acted upon by a force, it's motion will be with acceleration
4. The following four points may be noted at this stage:
(i) The force which pulls the block towards O is not a constant force
(ii) This is because, the force in the spring depends up on the distance from O
♦ Greater the distance, greater the force
(iii) While the motion is taking place, distance from O is decreasing
♦ So the force is decreasing
(iv) As a consequence, the acceleration is also decreasing
5. Thus we can say: The motion of the block towards O is under a 'varying acceleration'
• The magnitude of this acceleration decreases as the motion towards O proceeds
(Note that, a freely falling body is under a constant acceleration g)
■ Even though it is a 'decreasing acceleration', it is not a 'deceleration'
• If it was a deceleration, it would be acting opposite to the direction of motion
• But here it is acting in the same direction of motion
• This siuation is just like 'slowly releasing the accelerator pedal of a moving car'
• As the pedal is being released, the magnitude of the acceleration decreases.
• But even when the pedal is being released slowly, the acceleration is there
♦ The acceleration will become zero only when the pedal is completely released
• Once the accelerator pedal is completely released, and pressure is shifted to the brake pedal, deceleration begins
[Also note that, we are assuming the block to be on a smooth friction-less horizontal surface. So there is no friction to overcome. That means., there is no deceleration at all]
6. Because of the acceleration, the velocity of the block will go on increasing
• Let the velocity be 'vO' when the block reaches O
■ Then kinetic energy of the block at O will be $\mathbf\small{\frac{m\,v_O^2}{2}}$
7. So we have the following situation:
(i) The block was initially at A, a distance of xt from O
• At that position:
♦ Potential energy = $\mathbf\small{\frac{k\,x_t^2}{2}}$
♦ Kinetic energy = 0
(ii) Now the block is at O
• At this position:
♦ Potential energy = 0
♦ Kinetic energy = $\mathbf\small{\frac{m\,v_O^2}{2}}$
8. We can prove that $\mathbf\small{\frac{m\,v_O^2}{2}=\frac{k\,x_t^2}{2}}$
• For that, we use the work-energy theorem (Details here)
(i) Consider the motion of the block from A to O
• According to the work-energy theorem, we have:
KO - KA = Work done on the block
(ii) Let us write the values:
• KO = $\mathbf\small{\frac{m\,v_O^2}{2}}$
• KA = 0
• Work done on the block = Area of the red triangle = $\mathbf\small{\frac{k\,x_t^2}{2}}$
(iii) Substituting the values, we get:
$\mathbf\small{\frac{m\,v_O^2}{2}-0=\frac{k\,x_t^2}{2}}$
• Thus we get: $\mathbf\small{\frac{m\,v_O^2}{2}=\frac{k\,x_t^2}{2}}$
9. Let us write a summary of our findings so far. It can be written in 4 points:
(i) At A:
Potential energy = $\mathbf\small{\frac{k\,x_t^2}{2}}$
Kinetic energy = 0
(ii) At O:
Potential energy = 0
Kinetic energy = $\mathbf\small{\frac{m\,v_O^2}{2}}$
(iii) Potential energy at A = Kinetic energy at O
(iv) Based on the above three points, we can obtain this information:
• At O there is no potential energy. Only kinetic energy is available
♦ That available kinetic energy is equal to the potential energy at A
• Also there is no kinetic energy at A
■ So it is obvious that, all the potential energy at A has been converted into kinetic energy at O
Nothing was added. Nothing was lost
10. Now we can continue the discussion:
• The block is now at O. It has traveled from A to reach O
• Can it continue in the same direction?
11. Once it crosses O, the spring will begin to apply resistance. As if 'O' is the border point and the spring does not want the block to come in
• In spite of that resistance, the block will cross O and continue it's travel
• This is because, it has kinetic energy to overcome the resistance of the spring
12. How far will the block travel after crossing O?
• The answer is: It will continue to travel until all it's kinetic energy is used up
13. How is that kinetic energy used up?
Answer: It is used up for overcoming the resistance offered by the spring
14. The above steps (12) and (13) gives us a method to calculate the 'distance traveled after crossing O'
• Let 'xc' be this distance. It is shown in fig.6.33(c) above. After travelling distance '', the block will reach B
• The 'work done by the spring on the block' while it travels a distance 'xc'
= 'Area enclosed by the force-displacement graph' for the distance 'xc'
= $\mathbf\small{\frac{k\,x_c^2}{2}}$
• This area is shown in magenta color in fig.6.33(d)
15. This work must be equal to the kinetic energy available to the block when it is at O
• So we can write: $\mathbf\small{\frac{k\,x_c^2}{2}=\frac{m\,v_O^2}{2}}$
16. But from 8(iii), we have:
Kinetic energy at O = Potential energy at A = $\mathbf\small{\frac{k\,x_t^2}{2}}$
• So, in step (15), instead of $\mathbf\small{\frac{m\,v_O^2}{2}}$, we can put $\mathbf\small{\frac{k\,x_t^2}{2}}$
• So we get: $\mathbf\small{\frac{k\,x_c^2}{2}=\frac{k\,x_t^2}{2}}$
• Thus we get xt = xc
17. So distances xt and xc are the same
■ If we pull the block by a distance 'x' from the equilibrium position O, the block will go in the same 'x' beyond O.
18. So now the block is at B. It has lost all the kinetic energy
• But the lost kinetic energy is stored up as potential energy
• Because of this potential energy, the spring will push the block away towards O
(Just as it was pulled towards O from A)
• This cyclic process continues and the block will oscillate about O
19. This oscillation must continue indefinitely
• But in the real world, there will be friction between the block and the horizontal surface.
• There will also be air resistance.
• Because of that, a portion of the energy will be lost in each travel
• 'xc' will be less than 'xt' even in the travel from A to B
♦ In the return travel from B to A, 'xt' will be less than 'xc'
• 'xt' and 'xc' will progressively decrease with each cycle. So the oscillation will stop after some time
■ If there is no friction or air resistance, we will see a neat display of 'conservation of energy'
1. Consider a spring stretched by xt as shown in fig.6.32(a) below
 |
| Fig.6.32 |
Work required = Area of the red triangle
2. In some cases, we may want to know 'intermediate works'. This can be explained as follows:
• In fig.a, two points xi and xf are marked on the x axis
• We want to know how much work is required for stretching the spring from xi to xf
3. Obviously, this work is equal to the area of the shaded trapezium shown in fig.b
• The area of that trapezium can be calculated easily. All we need are the following 3 items:
(i) Height of the trapezium at xi
♦ This is equal to k xi
(ii) Height of the trapezium at xf
♦ This is equal to k xf
(iii) Base of the trapezium = (xf-xi)
• With the above 3 items, we will get the area. (Details about area of trapezium can be seen here)
■ So the work done will depend on the initial and final positions (xi and xf) only
4. What if, the path contains two segments as follows:
(i) The mass is moved from xi to xf
(ii) The mass is moved from xf to xi
• This the path is a cyclic path. The work done in segment-1 will be released in segment-2
■ Then the net work will be zero
5. So spring force is indeed a conservative force
• Since it is a conservative force, the energy stored in it (the spring potential energy) will follow the law of conservation of energy.
• So we can get back 'all the work done' at a later time
Now we will put this information into a practical application:
1. In fig.6.33(a) below, a block is pulled to a point A. It is kept at that position
 |
| Fig.6.33 |
• So a potential energy of $\mathbf\small{\frac{k\,x_t^2}{2}}$ will be stored in the block
2. At that position, the block is stationary. So it's kinetic energy K is zero
• If we release the block, the spring will pull it towards the equilibrium position 'O'
• We can say this:
When released, the block moves towards 'O'
3. During this motion, the block is acted upon by the force of the spring
(This is just like a freely falling body acted upon by gravitational force)
• Since the block is acted upon by a force, it's motion will be with acceleration
4. The following four points may be noted at this stage:
(i) The force which pulls the block towards O is not a constant force
(ii) This is because, the force in the spring depends up on the distance from O
♦ Greater the distance, greater the force
(iii) While the motion is taking place, distance from O is decreasing
♦ So the force is decreasing
(iv) As a consequence, the acceleration is also decreasing
5. Thus we can say: The motion of the block towards O is under a 'varying acceleration'
• The magnitude of this acceleration decreases as the motion towards O proceeds
(Note that, a freely falling body is under a constant acceleration g)
■ Even though it is a 'decreasing acceleration', it is not a 'deceleration'
• If it was a deceleration, it would be acting opposite to the direction of motion
• But here it is acting in the same direction of motion
• This siuation is just like 'slowly releasing the accelerator pedal of a moving car'
• As the pedal is being released, the magnitude of the acceleration decreases.
• But even when the pedal is being released slowly, the acceleration is there
♦ The acceleration will become zero only when the pedal is completely released
• Once the accelerator pedal is completely released, and pressure is shifted to the brake pedal, deceleration begins
[Also note that, we are assuming the block to be on a smooth friction-less horizontal surface. So there is no friction to overcome. That means., there is no deceleration at all]
6. Because of the acceleration, the velocity of the block will go on increasing
• Let the velocity be 'vO' when the block reaches O
■ Then kinetic energy of the block at O will be $\mathbf\small{\frac{m\,v_O^2}{2}}$
7. So we have the following situation:
(i) The block was initially at A, a distance of xt from O
• At that position:
♦ Potential energy = $\mathbf\small{\frac{k\,x_t^2}{2}}$
♦ Kinetic energy = 0
(ii) Now the block is at O
• At this position:
♦ Potential energy = 0
♦ Kinetic energy = $\mathbf\small{\frac{m\,v_O^2}{2}}$
8. We can prove that $\mathbf\small{\frac{m\,v_O^2}{2}=\frac{k\,x_t^2}{2}}$
• For that, we use the work-energy theorem (Details here)
(i) Consider the motion of the block from A to O
• According to the work-energy theorem, we have:
KO - KA = Work done on the block
(ii) Let us write the values:
• KO = $\mathbf\small{\frac{m\,v_O^2}{2}}$
• KA = 0
• Work done on the block = Area of the red triangle = $\mathbf\small{\frac{k\,x_t^2}{2}}$
(iii) Substituting the values, we get:
$\mathbf\small{\frac{m\,v_O^2}{2}-0=\frac{k\,x_t^2}{2}}$
• Thus we get: $\mathbf\small{\frac{m\,v_O^2}{2}=\frac{k\,x_t^2}{2}}$
9. Let us write a summary of our findings so far. It can be written in 4 points:
(i) At A:
Potential energy = $\mathbf\small{\frac{k\,x_t^2}{2}}$
Kinetic energy = 0
(ii) At O:
Potential energy = 0
Kinetic energy = $\mathbf\small{\frac{m\,v_O^2}{2}}$
(iii) Potential energy at A = Kinetic energy at O
(iv) Based on the above three points, we can obtain this information:
• At O there is no potential energy. Only kinetic energy is available
♦ That available kinetic energy is equal to the potential energy at A
• Also there is no kinetic energy at A
■ So it is obvious that, all the potential energy at A has been converted into kinetic energy at O
Nothing was added. Nothing was lost
10. Now we can continue the discussion:
• The block is now at O. It has traveled from A to reach O
• Can it continue in the same direction?
11. Once it crosses O, the spring will begin to apply resistance. As if 'O' is the border point and the spring does not want the block to come in
• In spite of that resistance, the block will cross O and continue it's travel
• This is because, it has kinetic energy to overcome the resistance of the spring
12. How far will the block travel after crossing O?
• The answer is: It will continue to travel until all it's kinetic energy is used up
13. How is that kinetic energy used up?
Answer: It is used up for overcoming the resistance offered by the spring
14. The above steps (12) and (13) gives us a method to calculate the 'distance traveled after crossing O'
• Let 'xc' be this distance. It is shown in fig.6.33(c) above. After travelling distance '', the block will reach B
• The 'work done by the spring on the block' while it travels a distance 'xc'
= 'Area enclosed by the force-displacement graph' for the distance 'xc'
= $\mathbf\small{\frac{k\,x_c^2}{2}}$
• This area is shown in magenta color in fig.6.33(d)
15. This work must be equal to the kinetic energy available to the block when it is at O
• So we can write: $\mathbf\small{\frac{k\,x_c^2}{2}=\frac{m\,v_O^2}{2}}$
16. But from 8(iii), we have:
Kinetic energy at O = Potential energy at A = $\mathbf\small{\frac{k\,x_t^2}{2}}$
• So, in step (15), instead of $\mathbf\small{\frac{m\,v_O^2}{2}}$, we can put $\mathbf\small{\frac{k\,x_t^2}{2}}$
• So we get: $\mathbf\small{\frac{k\,x_c^2}{2}=\frac{k\,x_t^2}{2}}$
• Thus we get xt = xc
17. So distances xt and xc are the same
■ If we pull the block by a distance 'x' from the equilibrium position O, the block will go in the same 'x' beyond O.
18. So now the block is at B. It has lost all the kinetic energy
• But the lost kinetic energy is stored up as potential energy
• Because of this potential energy, the spring will push the block away towards O
(Just as it was pulled towards O from A)
• This cyclic process continues and the block will oscillate about O
19. This oscillation must continue indefinitely
• But in the real world, there will be friction between the block and the horizontal surface.
• There will also be air resistance.
• Because of that, a portion of the energy will be lost in each travel
• 'xc' will be less than 'xt' even in the travel from A to B
♦ In the return travel from B to A, 'xt' will be less than 'xc'
• 'xt' and 'xc' will progressively decrease with each cycle. So the oscillation will stop after some time
■ If there is no friction or air resistance, we will see a neat display of 'conservation of energy'
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