Why is only half energy stored in an extended wire

In summary, the discussion relates to the extension of a uniform cross-section, homogenous, ideal wire within the proportional limit (Hooke's law). It is observed that only half of the gravitational potential energy lost when a mass is lowered on the initially un-extended wire is stored in the wire itself, while the remaining half remains in the mass. This is due to the stress-strain relation and can be further explained through a free body diagram and Newton's second law. The equation for the potential energy change, work done on the hand, and stored energy are all equivalent, with the stored energy being the sum of the other two energies.
  • #1
kirakun
25
2
This relates to the extension of a uniform cross-section, homogenous, ideal wire which extends within the proportional limit (Hooke's law).

From my understanding, only half of the gravitational potential energy lost when the mass is lowered on the initially un-extended wire, is stored in the wire itself as potential energy which increases the separation of molecules internally as long as the load is maintained.

According to my understanding, the remaining half of GPE in fact remains in the mass and this is eventually dissipated as heat (if there is surrounding air to oppose motion) and work is done on the hand which lowers the mass.

Why is only half but not all of the GPE stored in the wire? I was told that this is due to the stress-strain relation (straight line through origin) but its not that convincing.

Thank you.
 
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  • #2
Can you do a free body diagram on the mass, and write down Newton's second law on the mass? This will get us started. (Include the force of the hand, but temporarily leave out the air resistance).

Chet
 
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  • #3
Chestermiller said:
Can you do a free body diagram on the mass, and write down Newton's second law on the mass? This will get us started. (Include the force of the hand, but temporarily leave out the air resistance).

Chet

There would be 3 forces:

1. Force exerted by the wire on the mass (upwards), T
2. Force exerted by the hand on the mass (upwards), F
3. And the weight (downwards), mg

Resultant force = ma

The equation should be this:

mg - T - F = ma ?
 
  • #4
kirakun said:
There would be 3 forces:

1. Force exerted by the wire on the mass (upwards), T
2. Force exerted by the hand on the mass (upwards), F
3. And the weight (downwards), mg

Resultant force = ma

The equation should be this:

mg - T - F = ma ?
This is excellent. Now, we are going to lower the weight gradually, so that F = mg - T. If the spring is at its unextended length to begin with, then, as you lower the weight, T = ky, where y is how much you lower it. Initially, y = 0, T =0, and F = mg. In the final position, F = 0, T = mg, and y = mg/k. How much work was done on your hand by the mass when you lowered the weight gradually? How much energy was stored in the spring? What was the change in potential energy of the mass?

Chet
 
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  • #5
Chestermiller said:
This is excellent. Now, we are going to lower the weight gradually, so that F = mg - T. If the spring is at its unextended length to begin with, then, as you lower the weight, T = ky, where y is how much you lower it. Initially, y = 0, T =0, and F = mg. In the final position, F = 0, T = mg, and y = mg/k. How much work was done on your hand by the mass when you lowered the weight gradually? How much energy was stored in the spring? What was the change in potential energy of the mass?

Chet

Hmm the answer seems right there but its not clicking. Any more hints.

The work done on the hand by mass = average force x distance = (mg/2 x mg/k) ?
Stored energy = Tension x distance = (mg x mg/k) ?

(Which is obviously not good since we know beforehand that the stored energy = work done on hand)Edit: Oh wait is it that the tension increases linearly with the distance moved such that the average force = (mg/2) and the stored energy becomes (mg/2 x mg/k) = work done on hand?

And the change in potential energy = mgh = (mg x mg/k) which is the sum of the other two energies.
 
Last edited:
  • #6
kirakun said:
Hmm the answer seems right there but its not clicking. Any more hints.

The work done on the hand by mass = average force x distance = (mg/2 x mg/k) ?
Stored energy = Tension x distance = (mg x mg/k) ?

(Which is obviously not good since we know beforehand that the stored energy = work done on hand)


Edit: Oh wait is it that the tension increases linearly with the distance moved such that the average force = (mg/2) and the stored energy becomes (mg/2 x mg/k) = work done on hand?

And the change in potential energy = mgh = (mg x mg/k) which is the sum of the other two energies.

Yes. Yes. Yes.

Writing your result a little differently,

change in potential energy = mgh

work done on hand = mgh/2

change in stored energy = mgh/2

This is what you were trying to show, right?

Chet
 
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  • #7
Chestermiller said:
Yes. Yes. Yes.

Writing your result a little differently,

change in potential energy = mgh

work done on hand = mgh/2

change in stored energy = mgh/2

This is what you were trying to show, right?

Chet

Thank you!
 

1. Why is only half energy stored in an extended wire?

There are several factors that contribute to only half of the energy being stored in an extended wire. One reason is that as the wire is extended, its resistance increases, leading to a decrease in the flow of current and therefore, a decrease in the stored energy. Additionally, as the wire is extended, some of the energy is lost as heat due to the increased resistance. Lastly, the wire's capacitance also plays a role, as it decreases with increased length, resulting in a decrease in stored energy.

2. How is the energy stored in an extended wire calculated?

The energy stored in an extended wire can be calculated using the formula E = 0.5 * L * I^2, where E is the energy in joules, L is the inductance in henries, and I is the current in amperes. This formula takes into account the effects of resistance, capacitance, and inductance on the stored energy.

3. Can the energy stored in an extended wire be increased?

Yes, the energy stored in an extended wire can be increased by decreasing its resistance, increasing its inductance, or increasing the current flowing through it. This can be achieved by using materials with lower resistance, increasing the number of turns in the wire to increase its inductance, or using a higher voltage source to increase the current.

4. How does the length of the wire affect the stored energy?

The length of the wire directly affects the stored energy, as it is one of the factors in the formula for calculating the energy. As the wire is extended, its inductance and capacitance change, leading to a decrease in the stored energy. Additionally, the longer the wire, the higher its resistance, resulting in more energy being lost as heat.

5. Does the type of material used in the wire affect the stored energy?

Yes, the type of material used in the wire can affect the stored energy. Different materials have different resistivity, which can impact the wire's resistance and therefore, the stored energy. Materials with lower resistivity, such as copper, will have less energy loss and therefore, a higher energy storage capacity compared to materials with higher resistivity.

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