# Why is only half energy stored in an extended wire

1. Apr 7, 2014

### kirakun

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.

2. Apr 7, 2014

### Staff: Mentor

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

3. Apr 7, 2014

### kirakun

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. Apr 7, 2014

### Staff: Mentor

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

5. Apr 7, 2014

### kirakun

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: Apr 7, 2014
6. Apr 7, 2014

### Staff: Mentor

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

7. Apr 7, 2014

Thank you!