Relationship between diameter and elastic potential energy of a wire

AI Thread Summary
The discussion centers on the relationship between the diameter of a wire and its elastic potential energy. It is proposed that doubling the diameter results in the elastic potential energy being reduced to 1/16th of its original value due to changes in cross-sectional area and stress. However, others point out that the correct reduction is actually to a quarter of the original value. There is confusion regarding the spring constant 'k' and its relationship to the cross-sectional area, with suggestions that 'k' may change as the area changes. The conversation emphasizes the importance of understanding the formulas related to stress, strain, and energy in elastic materials.
RateOfReturn
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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 ?
 
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RateOfReturn said:
2) F= stress/area.
Are you sure?
 
Steve4Physics said:
Are you sure?
Sorry, it was a typo- I meant to say stress = force/area. I'll edit the original question.
 
RateOfReturn said:
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.
 
Steve4Physics said:
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 ?
 
Is the force in the formula Stress= Force/area and the force in Force= spring constant times extension, the same force ?
 
RateOfReturn said:
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.
 
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RateOfReturn said:
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).
 
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