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Hookes Law and a uniform bar

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A uniform bar of an iron is supported by a long, uniform hooke's law spring. The spring is cut in half and two pieces are used to support the same bar. If the whole spring stretched by 4.0cm, by how much would each half strech?

    2. Relevant equations
    f=Kchange in X

    3. The attempt at a solution
  2. jcsd
  3. Apr 30, 2007 #2

    Doc Al

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    Staff: Mentor

    Hint: How does the spring constant of each half-spring compare to the original spring constant? (Does it get harder or easier to stretch?)
  4. Apr 30, 2007 #3
    it become harder to stretch...
  5. Apr 30, 2007 #4
    I tried solving it and dats what i got:

    Fnet= ma
    Fs + fg = 0
    KX - mg = 0
    k = (mg)/ X
    (10kg x 9.8m/s) / 0.04
    = 2450

    first spring = (2450 x 2)
    = 4 900
    Fa = (k1 + k2)
    X= (mg/ k1 + k2)
    (10kg x 9.8) / ( 4 900 N/m+4 900N/m)
    0.01m= X
  6. Apr 30, 2007 #5
    please let me know if i did it the right way or not...Thank You so Much!!
  7. Apr 30, 2007 #6

    Doc Al

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    Staff: Mentor

    Looks good to me! You assumed a mass of 10 Kg for the iron bar (which was not given); but since the answer does not depend on the mass, that's an OK strategy.

    But realize you can also solve this algebraically without assuming values:

    For the whole spring:
    W = KX (where W is the weight of the bar and X = 4 cm)

    So K = W/X

    For the half-springs:
    k = 2K = 2W/X

    Since each spring supports half the weight:
    W/2 = kx = (2W/X)x

    So x = X/4 = 1 cm
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