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Energy stored in the wire

  1. Jan 8, 2015 #1
    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. jcsd
  3. Jan 8, 2015 #2

    CWatters

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


     
  4. Jan 8, 2015 #3
    It is applicable to any specimens when they are extended or compressed within their limits.
     
  5. Jan 8, 2015 #4

    BvU

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    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 ?)
     
  6. Jan 8, 2015 #5
    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.
     
  7. Jan 8, 2015 #6

    BvU

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    Well, that IS average force times extension
     
  8. Jan 8, 2015 #7
    But I just don't understand why i was wrong. Could you please explain how they get the result physically?
     
  9. Jan 8, 2015 #8
    Anybody explain the result for me please ?
     
  10. Jan 8, 2015 #9
    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 ,,
    :)
     
  11. Jan 8, 2015 #10
    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 ??
     
  12. Jan 8, 2015 #11

    BvU

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    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

    [edit] the integral bounds look a bit stupid. lower bound is equilibrium position (T=0)
     
    Last edited: Jan 8, 2015
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