Relationship between diameter and elastic potential energy of a wire

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Homework Statement:

A wire is replaced by a different wire with the same material, length but double the diameter. The same load is attached to the wire. What is the change to the elastic potential energy

Relevant Equations:

F=kx
E= 1/2kx^2
Young's Modulus = Stress/strain

I think the answer is that the elastic potential energy will be a 1/16th of the original value. This is my reasoning:

1) If the diameter doubles, the cross sectional area is 4 times the original value. (from A= πr²).
2) F= stress/area. Force (load is the same). If cross sectional area quadruples, then the stress must quarter.
3) E= stress/strain. E is the same. If stress quarters, than strain must quarter. If strain quarters, extension must quarter (since original length is still the same).
4) From E= 1/2 kx², with k being the same and extension being a quarter of the original value, the energy must be a 1/16th of the original value.

However, the answers say the elastic potential energy is a quarter of the original value. Which part of my reasoning is incorrect ?

 

[Steve4Physics]

2) F= stress/area.

Are you sure?

 

[RateOfReturn]

Are you sure?

Sorry, it was a typo- I meant to say stress = force/area. I'll edit the original question.

 

[Steve4Physics]

Sorry, it was a typo- I meant to say stress = force/area. I'll edit the original question.

Yes. I made a typo' too - hit the send button by mistake. I had another comment which is:

In 1/2kx^2, will 'k' remain the same if the cross-sectional area is changed?

Edit. Ah, @TSny beat me to it.

 

[RateOfReturn]

Yes. I made a typo' too - hit the send button by mistake. I had another comment which is:

In 1/2kx^2, will 'k' remain the same if the cross-sectional area is changed?

I am not familiar with how k and cross-sectional area are related. Although, I'm thinking if the cross-sectional area quadruples, then the force needed per unit extension must also quadruple. So k must quadruple ?

 

[RateOfReturn]

Is the force in the formula Stress= Force/area and the force in Force= spring constant times extension, the same force ?

 

[Steve4Physics]

I am not familiar with how k and cross-sectional area are related. Although, I'm thinking if the cross-sectional area quadruples, then the force needed per unit extension must also quadruple. So k must quadruple ?

If we use ‘E’ for the Young modulus and ‘e’ (rather than x) for extension, then we have the standard formula$$E = \frac {FL}{eA}$$or as my old physics teacher used to remind us: ‘Young has flea(s)".

You can use the formula to answer your original question directly. Or, if you want to express k in term of E, A and L, it’s not too hard if you rearrange the formula to make F the subject.

 

[Steve4Physics]

Is the force in the formula Stress= Force/area and the force in Force= spring constant times extension, the same force ?

Yes.

Edit. You might like to note that the energy stored is also given by the expression ½Fx. This corresponds to the average force during stretching (½F) times the extension (x).