# Relationship between diameter and elastic potential energy of a wire

• RateOfReturn
In summary, if the diameter of an object doubles, the cross-sectional area will increase by 4 times the original value. This will cause the stress to decrease to a quarter of the original value, leading to a quarter of the strain and extension as well. Therefore, according to the formula E=1/2kx^2, where k is the same and extension is a quarter of the original value, the elastic potential energy will be a 1/16th of the original value. However, it is possible that the force in the formula Stress= Force/area and the force in Force= spring constant times extension may not be the same force, which could cause discrepancies in the calculations.
RateOfReturn
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 ?

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.

Last edited:
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).

Last edited:

## 1. What is the relationship between the diameter of a wire and its elastic potential energy?

The diameter of a wire and its elastic potential energy have a direct relationship. This means that as the diameter of a wire increases, its elastic potential energy also increases. Similarly, as the diameter decreases, the elastic potential energy decreases.

## 2. How does the diameter of a wire affect its stiffness and elastic potential energy?

The diameter of a wire is directly proportional to its stiffness and elastic potential energy. A thicker wire will have a higher stiffness and elastic potential energy compared to a thinner wire with the same material and length.

## 3. Can the diameter of a wire affect its elastic limit?

Yes, the diameter of a wire can affect its elastic limit. A thicker wire will have a higher elastic limit compared to a thinner wire. This is because a thicker wire has a larger cross-sectional area, allowing it to withstand greater forces before reaching its elastic limit.

## 4. Why is the diameter of a wire important in determining its elastic potential energy?

The diameter of a wire is important in determining its elastic potential energy because it affects the wire's stiffness and elastic limit. A thicker wire will have a higher elastic potential energy compared to a thinner wire, making it more suitable for applications that require high levels of energy storage and release.

## 5. How does the material of a wire factor into the relationship between diameter and elastic potential energy?

The material of a wire plays a significant role in the relationship between diameter and elastic potential energy. Different materials have different stiffness and elastic limits, which can affect the wire's elastic potential energy. For example, a copper wire will have a higher elastic potential energy compared to a steel wire with the same diameter due to its higher stiffness.

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