Elastic and gravitational potential energy

[Pratik89]

X has a mass of 55 kg and hangs from a rope. As a result of this the rope stretches by 0.6 m. Calculate the energy stored in the rope as a result of stretching.
Solution:
F = kx and E = 0.5x^2. Using this fetches the answer as 161.5
However, the change in potential energy (calculated using mgh ) of X is exactly twice of this energy stored. Is there a relationship between the two?. What is the explanation

 

[QuantumQuest]

X has a mass of 55 kg and hangs from a rope. As a result of this the rope stretches by 0.6 m. Calculate the energy stored in the rope as a result of stretching.
Solution:
F = kx and E = 0.5x^2. Using this fetches the answer as 161.5
However, the change in potential energy (calculated using mgh ) of X is exactly twice of this energy stored. Is there a relationship between the two?. What is the explanation

What sort of things are taking place as the rope stretches and potential energy of the mass gets lower?

 

[Pratik89]

What sort of things are taking place as the rope stretches and potential energy of the mass gets lower?

That is the question I am trying to understand

 

[BvU]

Hello Pratik,

There is a relationship: Gravity does work to the tune of 323 J when the weight is lowered. One half ends up in the spring, the other half is delivered to the device (a hand, or whatever) that lowers X from its initial position to 0.6 m lower.

 

[Pratik89]

Hello Pratik,

There is a relationship: Gravity does work to the tune of 323 J when the weight is lowered. One half ends up in the spring, the other half is delivered to the device (a hand, or whatever) that lowers X from its initial position to 0.6 m lower.

But isn't the whole body being lowered and not just the hand ? The hand is part of the body so the energy loss has been accounted for already

 

[Pratik89]

Hello Pratik,

There is a relationship: Gravity does work to the tune of 323 J when the weight is lowered. One half ends up in the spring, the other half is delivered to the device (a hand, or whatever) that lowers X from its initial position to 0.6 m lower.

And one half ends up in the rope, the other in the hand/device, does not sound quite convincing

 

[haruspex]

But isn't the whole body being lowered and not just the hand ? The hand is part of the body so the energy loss has been accounted for already

That's not the hand BvU meant.
What would happen if you started with an unstretched rope, tied a mass to the end while keeping the rope taut but unstretched, then let go?

 

[Pratik89]

That's not the hand BvU meant.
What would happen if you started with an unstretched rope, tied a mass to the end while keeping the rope taut but unstretched, then let go?

It is not clear what you are trying to say. I will refine the question, "when a point mass is attached to a rope, the rope will stretch, why is the loss in gravitational potential energy of the mass NOT equal to the gain in elastic energy of rope?"

 

[haruspex]

It is not clear what you are trying to say

To make it clearer, consider a light spring. It hangs, unstretched. You carefully tie a mass to the end, not allowing the spring to become stretched as you do so. Now you let go. What do you expect to happen over the next little while?

 

[Pratik89]

The spring would stretch and the mass would fall, however, I don't see how this answers my question

 

[haruspex]

The spring would stretch and the mass would fall,

Is that it? What happens after the mass stops falling?

 

[Pratik89]

Is that it? What happens after the mass stops falling?

Can you put a simple explanation or answers instead of counter-questions

 

[conscience]

Hi Pratik ,

Can you put a simple explanation or answers instead of counter-questions

You have posed a nice conceptual doubt . Please do not lose patience . @haruspex is guiding you nicely . Give some thought to his questions and answer accordingly . You will surely benefit from this discussion .

 

[Pratik89]

Hi Pratik ,
You have posed a nice conceptual doubt . Please do not lose patience . @haruspex is guiding you nicely . Give some thought to his questions and answer accordingly . You will surely benefit from this discussion .

Hi,
Thank you for the response. I really fail to see how this will get me to the answer. I am a Physics teacher myself and am aware of the necessity of patience while understanding things. However, there are times when a simple/clear explanation is all a student wants instead of building it up painfully slowly with seemingly obscure questions. Instead of patronizing me, if you could elaborate on the point Haruspex is trying to get to, it would benefit all of us.

 

[haruspex]

Can you put a simple explanation or answers instead of counter-questions

Learning to ask the right questions is crucial. Please try to answer mine. It is not difficult, and it will lead you to the answer you seek.

 

[Pratik89]

Learning to ask the right questions is crucial. Please try to answer mine. It is not difficult, and it will lead you to the answer you seek.

Alright. The mass will drop initially, and then it will bounce back, the oscillations will continue for a while before they dampen and the mass settles down at a particular height. At this point, the mass has lost some energy (because of the change in height) and the the spring has gained some energy (because of the extension)

 

[haruspex]

Alright. The mass will drop initially, and then it will bounce back, the oscillations will continue for a while before they dampen and the mass settles down at a particular height. At this point, the mass has lost some energy (because of the change in height) and the the spring has gained some energy (because of the extension)

Ok, good.
Now, if it were perfectly elastic, instead of converging to the equilibrium position, it would oscillate forever. What are the relationships between:

• The position it was initially released from,
• The equilibrium position, and
• The lowest point of the oscillation

?
Then think about the energy differences, and what that tells you about the fraction of energy that is eventually lost in the imperfectly elastic case.

 

[Pratik89]

Ok, good.
Now, if it were perfectly elastic, instead of converging to the equilibrium position, it would oscillate forever. What are the relationships between:

• The position it was initially released from,
• The equilibrium position, and
• The lowest point of the oscillation

?
Then think about the energy differences, and what that tells you about the fraction of energy that is eventually lost in the imperfectly elastic case.

the equilibrium position should be in between the lowest point and the point it was initially released from. So are you saying that, the energy difference is because there is some loss of energy while the mass oscillates ?

 

[conscience]

the equilibrium position should be in between the lowest point and the point it was initially released from.

OK.

So are you saying that, the energy difference is because there is some loss of energy while the mass oscillates ?

No.

Pratik ,

There are two ways in which the mass could come down from initial unstretched position of spring.

1 . The support holding the mass is suddenly removed and the mass is allowed to fall such that the forces acting on it are spring force and force due to gravity .

2. The mass is very slowly allowed to come down such that it is in equilibrium at all times .The forces acting on the mass are spring force , force due to gravity and some external supporting force ( may be your hand ) .

Which of the two cases are we talking about in this question ?

 

[haruspex]

OK.
No.

Pratik ,

There are two ways in which the mass could come down from initial unstretched position of spring.

1 . The support holding the mass is suddenly removed and the mass is allowed to fall such that the forces acting on it are spring force and force due to gravity .

2. The mass is very slowly allowed to come down such that it is in equilibrium at all times .The forces acting on the mass are spring force , force due to gravity and some external supporting force ( may be your hand ) .

Which of the two cases are we talking about in this question ?

Just a moment...

I am not entirely sure what Pratik's main question is, whether it is where the energy goes, or why it is exactly half. The way I read the OP I thought it was mainly the second, so I have been concentrating on that.

For where the energy goes, there are two main possibilities. As you suggest, it might be all lost in how the mass is released, but if not, it will be energy lost during the oscillations. The question will then be, why is energy lost in oscillating.

 

[conscience]

I am not entirely sure what Pratik's main question is, whether it is where the energy goes, or why it is exactly half. The way I read the OP I thought it was mainly the second, so I have been concentrating on that.

Both . Why only half the loss in gravitational potential energy is stored in the rope and where does the other half go ?

but if not, it will be energy lost during the oscillations. The question will then be, why is energy lost in oscillating.

I do not think the question is talking about energy lost in oscillations.But I may be wrong .

 

[Pratik89]

Just a moment...

I am not entirely sure what Pratik's main question is, whether it is where the energy goes, or why it is exactly half. The way I read the OP I thought it was mainly the second, so I have been concentrating on that.

For where the energy goes, there are two main possibilities. As you suggest, it might be all lost in how the mass is released, but if not, it will be energy lost during the oscillations. The question will then be, why is energy lost in oscillating.

My question is why is it exactly half and where does the other half end up ?

 

[haruspex]

My question is why is it exactly half and where does the other half end up ?

Ok, but for the latter you need to answer the question both BvU and conscience asked: how exactly is the mass released? Is it let go from the unstretched position of the rope/spring, or is it lowered to the equilibrium position and then released? Or maybe something between these two?

 

[Pratik89]

Ok, but for the latter you need to answer the question both BvU and conscience asked: how exactly is the mass released? Is it let go from the unstretched position of the rope/spring, or is it lowered to the equilibrium position and then released? Or maybe something between these two?

The question does not specify this, but let us assume is it released slowly to the equilibrium position. In either case energy stored in the rope (due to stretching) should be the same

 

[haruspex]

The question does not specify this, but let us assume is it released slowly to the equilibrium position. In either case energy stored in the rope (due to stretching) should be the same

I switched to discussing springs because I feel it is more obvious what is going on in that context, and the result applies to ropes also.
If you attach a mass to an unstretched spring then lower it to the equilibrium position, what force acts on your hand as you do so? Does this force do work?

 

[Pratik89]

I switched to discussing springs because I feel it is more obvious what is going on in that context, and the result applies to ropes also.
If you attach a mass to an unstretched spring then lower it to the equilibrium position, what force acts on your hand as you do so? Does this force do work?

The force acting on the hand would be the force due to the mass of the object.

 

[Doc Al]

The question does not specify this, but let us assume is it released slowly to the equilibrium position.

That is key. So another force besides gravity must act, else the object wouldn't just "hang" there but would oscillate up and down.

In either case energy stored in the rope (due to stretching) should be the same

That's true.

The force acting on the hand would be the force due to the mass of the object.

Do you mean the weight of the object? How much upward force does the hand have to exert to "slowly" lower the mass to the stretched point? Does that force change as the object lowers?

 

[Pratik89]

That is key. So another force besides gravity must act, else the object wouldn't just "hang" there but would oscillate up and down.That's true.Do you mean the weight of the object? How much upward force does the hand have to exert to "slowly" lower the mass to the stretched point? Does that force change as the object lowers?

Well, yes the force changes, initially it is equal to the force exerted by the mass of the object, however as it lowers down, the force reduces and finally at equilibrium, the force becomes zero.

 

[Doc Al]

Well, yes the force changes, initially it is equal to the force exerted by the mass of the object, however as it lowers down, the force reduces and finally at equilibrium, the force becomes zero.

Exactly. Does that force do work on the system? How much work, compared to the change in gravitational PE?

 

[Pratik89]

Exactly. Does that force do work on the system? How much work, compared to the change in gravitational PE?

It reduces linearly, so the plot of force versus distance would be a dropping line, the energy would be the area under this which is 0.5F*x.

 

[Doc Al]

It reduces linearly, so the plot of force versus distance would be a dropping line, the energy would be the area under this which is 0.5F*x.

Good. (Where F is the weight and x the full stretch.) Is that work positive or negative? And how does it compare to the work done by gravity?

 

[Pratik89]

Good. (Where F is the weight and x the full stretch.) Is that work positive or negative? And how does it compare to the work done by gravity?

it should be positive because the direction of force and motion is the same and I think it would be half of the work done by gravity

 

[haruspex]

it should be positive because the direction of force and motion is the same and I think it would be half of the work done by gravity

Right, but how did you decide it was half?

 

[lychette]

My question is why is it exactly half and where does the other half end up ?

It is only half if force is proportional to extension (straight line graph of F against x )

 

[Doc Al]

it should be positive because the direction of force and motion is the same

What's the direction of the force exerted by the hand? What's the direction of motion?