Why is only half energy stored in an extended wire

  • Context: Undergrad 
  • Thread starter Thread starter kirakun
  • Start date Start date
  • Tags Tags
    Energy Wire
Click For Summary

Discussion Overview

The discussion revolves around the energy storage in an extended wire when a mass is lowered, specifically addressing why only half of the gravitational potential energy (GPE) is stored in the wire itself while the other half remains with the mass. The conversation touches on concepts related to Hooke's law, energy transfer, and the mechanics involved in lowering the mass.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant notes that when a mass is lowered on an initially un-extended wire, only half of the GPE lost is stored in the wire, while the other half dissipates as heat or work done on the hand.
  • Another participant suggests creating a free body diagram and applying Newton's second law to analyze the forces acting on the mass, including the tension in the wire and the force exerted by the hand.
  • There is a discussion about the relationship between tension and the distance lowered, with references to the linear increase of tension as the mass is lowered.
  • Participants explore the calculations for work done on the hand, stored energy in the wire, and changes in potential energy, with some expressing confusion over the relationships between these quantities.
  • One participant proposes that the average force exerted during the lowering process is half of the weight of the mass, leading to a conclusion about the work done and energy stored.

Areas of Agreement / Disagreement

While there is a general understanding that only half of the GPE is stored in the wire, the discussion includes multiple viewpoints on the mechanics and calculations involved, and no consensus is reached on the underlying reasons for this phenomenon.

Contextual Notes

Participants express uncertainty regarding the calculations and relationships between the energies involved, indicating a dependence on assumptions about the linearity of tension and the definitions of work and energy in this context.

kirakun
Messages
24
Reaction score
1
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.
 
Physics news on Phys.org
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
 
  • Like
Likes   Reactions: 1 person
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 ?
 
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
 
  • Like
Likes   Reactions: 1 person
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:
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
 
  • Like
Likes   Reactions: 1 person
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!
 

Similar threads

Undergrad I would like a better understanding of friction and hysteresis
  • · Replies 5 ·
Replies
5
Views
3K
Graduate Understanding the derivation for Elastic Potential Energy.
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad How Does Gravity Affect Potential Energy in Physics?
  • · Replies 18 ·
Replies
18
Views
3K
Undergrad Question about Momentum vs. Kinetic Energy vs. Deforming In Collisions
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Same old question, I still don't get it: E=1/2mv^2 - more E with V?
  • · Replies 41 ·
2
Replies
41
Views
4K
Graduate Conservative forces, Nonconservative forces and Potential Energy
  • · Replies 9 ·
Replies
9
Views
4K
Undergrad How to understand experiment to measure heat capacity of solid?
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Why is there close to zero potential difference in conductors in a circuit?
  • · Replies 21 ·
Replies
21
Views
10K
Why Does the Filament Box in a CRT Electron Gun Have Multiple Pins?
  • · Replies 12 ·
Replies
12
Views
2K
Undergrad Mechanism of Energy Conservation in Zero-Amplitude Sum of EM Waveforms
  • · Replies 21 ·
Replies
21
Views
3K