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Strain Energy in a Spring

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data
    A spring hangs vertically from a fixed point. When a mass of 94g is suspended from the spring the spring stretches by 2.6 cm before temporarily coming to rest. Determine the extension of the spring at which the strain energy is half its maximum value.


    2. Relevant equations

    F=0.5kx^2

    3. The attempt at a solution
    Loss in Gravitational Potential Energy = Gain in Strain Energy
    So Half Strain energy = Half Gravitational Potential Energy
    0.5k(.026*10^-2)^2=0.012 so k=6000/169
    and since k=F/x using F=0.094*9.8 to get x
     
  2. jcsd
  3. May 6, 2013 #2

    haruspex

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    I don't understand your working there. Pls post full details. .026*10^-2 doesn't look right - do you mean .026? For k, I get about 70N/m.
     
  4. May 6, 2013 #3
    Oops sorry. Yes I meant 0.026 but my value for k is 6000/169 (approx. 35.5)
    Then I used the Weight force and my k in F=kx (hook's law) to get x so (0.094*9.8)/(35.5)=x
     
  5. May 6, 2013 #4

    haruspex

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    OK, but I asked you to explain how you got that equation. Where does the .012 come from? That looks to me like half the strain energy at max extension, but to compute k you need to be using the whole strain energy at max extension.
     
  6. May 7, 2013 #5
    edited post
     
    Last edited: May 7, 2013
  7. May 7, 2013 #6
    Yes the .012 is half the gravitational potential energy lost (mgh) which is equal to half the strain energy gained(1/2 k x^2) (Using conservation of energy)
    So I wrongly equated the whole strain energy to half the strain energy. I corrected it now:
    .25k(.026)^2=.012
    Now I get k=71
    Is that how you got it ?
     
  8. May 7, 2013 #7

    haruspex

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    No. You are told:
    That is enough information to calculate k. No mention is made of any half amounts of energy yet.
     
  9. May 7, 2013 #8
    O.K but in the beginning I tried using F=ke to get k. for F= .094*9.8 and e=.026, k=35.4
    Why didn't I get the true value for k ?
     
  10. May 7, 2013 #9

    haruspex

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    Because that equation assumes the extension is to the equilibrium point (zero acceleration), but the extension given is to zero KE point.
     
  11. May 8, 2013 #10
    Thank you so much :D
     
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