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Young's Modulus and Hooke's Law

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A wire of length L, Young's modulus Y, and cross-sectional area A is stretched
    elastically by an amount ∆L. The restoring force is given by Hooke's Law as
    k∆L.

    b. show that the work done in stretching the wire must be: W = (YA/2L) x (∆L)^2.




    3. The attempt at a solution

    Not sure what work equation I use and how to put it into the equation. I tried
    W = F x ∆L but my answer ended up being W = (∆L)^2 x YA x L
     
    Last edited: Jun 16, 2010
  2. jcsd
  3. Jun 16, 2010 #2

    PhanthomJay

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    The restoring force F is not constant over the stretched distance, it varies linearly from 0 to the maximum value of k(delta L). You can use calculus to calculate the work done, or just use the average force in your work formula. What's the value of k in terms of A, E, and L?
     
  4. Jun 16, 2010 #3
    1)From the Young's Modulus equation: ∆L = (1/Y) (F/A) x L = (1/Y) (k∆L)/A x L

    So I isolated for k and get, k = YA/L

    2) I then isolated for k from the work equation: W = (k∆L)∆L

    ...which gave me k = W/(∆L^2)

    So when I make the 2 equations equal to eachother, I see that I'm almost there, but where does the "2" in 2L come from? That's all I am missing
     
  5. Jun 16, 2010 #4

    PhanthomJay

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    You missed my tip: the work is not k(delta L)^2, because the force , k(delta x), is not constant. Use the average force.
     
  6. Jun 16, 2010 #5
    Hmmmm...I don't know why I can't think what the average force is? Is it F = k∆L/2 ? Stumped
     
  7. Jun 16, 2010 #6

    PhanthomJay

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    Yes, if the force varies from 0 as it just starts to stretch, to k(deltaL) when it is stretched to deltaL, then the average force is the (sum of the min force plus maxforce)/2. Or you can use the calculus for the work done, the integral of F(dx), where F=kx.
     
  8. Jun 16, 2010 #7
    Awesome, it all worked out and I got the right answer. Thank you so much!
     
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