# Energy stored in the wire

## Homework Statement

Under tension T the wire has lenghth of m, its length becomes n when the tension is increase to T'. What is the extra energy stored in the wire as a result of this process?

E=1/2 Fe[/B]

## The Attempt at a Solution

Extra energy= E(n)-E(m) = 1/2 x (T'n-Tm)

There is no such answer in the multiple choices. Am I wrong when I subtracted the two energy?[/B]

CWatters
Homework Helper
Gold Member
I might be wrong but is that the right equation for the energy stored in a "spring"?

I might be wrong but is that the right equation for the energy stored in a "spring"?

It is applicable to any specimens when they are extended or compressed within their limits.

BvU
Homework Helper
E=1/2 Fe looks like some average force times extension when starting at zero force. $$\int_n^m T\; ds$$ would be a lot better. See what that gives. (What are the multiple choice options to choose from ?)

I don't think they require the exact energy. The right answer is 1/2(T'+T)(n-m) but I have no idea why.

BvU
Homework Helper
Well, that IS average force times extension

But I just don't understand why i was wrong. Could you please explain how they get the result physically?

Anybody explain the result for me please ?

1/2(T'+T)(n-m)

If I'm not mistaken ,, I think this is related a work-energy theorem
the energy is equal to the external work

So
$$E_f-E_i = 0.5 (T'_n-T_m)(n-m)$$

the left side is the energy difference , the right side is the average work done by the tension force [average force times the distance]
this is what i can say ,,
:)

Extra energy= E(n)-E(m) = 1/2 x (T'n-Tm)

I think that is not the right answer ,,

the left side is an energy , while the right side is a force ,,,

Check the units of each side ,,,

Are they equal ??

BvU
Homework Helper
the left side is an energy , while the right side is a force
I thought the same thought at first, until I realized it (the n and the m) was a multiplication, not a subscript.

The formula stems from inserting T = k x and ds = dx in $$\int_0^n T\; ds = \int_0^n kx\; dx = \left [{1\over 2} k x^2 \right ]^n_0 = {1\over 2} kx\; x = {1\over 2} Tx$$

(I strongly prefer and recommend ## {1\over 2} kx^2 ## !!!)

Numerically there isn't much difference between the 'absent' and the 'right' multiple choice

 the integral bounds look a bit stupid. lower bound is equilibrium position (T=0)

Last edited: