Energy stored in the wire

1. Jan 8, 2015

Tulatalu

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
E=1/2 Fe

3. 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?

2. Jan 8, 2015

CWatters

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

3. Jan 8, 2015

Tulatalu

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

4. Jan 8, 2015

BvU

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 ?)

5. Jan 8, 2015

Tulatalu

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.

6. Jan 8, 2015

BvU

Well, that IS average force times extension

7. Jan 8, 2015

Tulatalu

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

8. Jan 8, 2015

Tulatalu

Anybody explain the result for me please ?

9. Jan 8, 2015

Maged Saeed

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 ,,
:)

10. Jan 8, 2015

Maged Saeed

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 ??

11. Jan 8, 2015

BvU

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: Jan 8, 2015