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How much work must you do to change length of spring?

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A spring has a relaxed length of 7 cm and a stiffness of 200 N/m. How much work must you do to change its length from 10 cm to 14 cm?

    2. Relevant equations
    Ef = Ei + W

    3. The attempt at a solution
    Ef = Ei + W
    Ef = Ei + F*delta x
    F*delta x = Ef - Ei
    F*delta x = (Ki + Ui) - (Kf + Uf)
    F*delta x = (1/2mvi^2 + 1/2ks si^2) - (1/2mvf^2 + 1/2ks sf^2)

    I'm trying to break this equation apart to try to figure this out, but I'm not sure what to do next. Help!

    Thanks
     
  2. jcsd
  3. Mar 1, 2009 #2

    LowlyPion

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    You're making things too hard.

    W = ∫ F*X dx

    And what is F?

    F = -k*x

    so ...
     
  4. Mar 1, 2009 #3
    Thinking about Hooke's Law

    Since,

    Force = -k * delta x

    Force = 200N/m * 14 - 10 cm

    or


    Force = 200N/m * 0.04m

    = Force required = 8N

    Work = Force x Distance

    Work = 8N x .04m

    = 0.32 J
     
  5. Mar 1, 2009 #4
    Wouldn't it be -0.32 J since F = -k * delta x?
     
  6. Mar 1, 2009 #5
    Sure =)


    I ignored the -...
     
  7. Mar 1, 2009 #6
    Hmm well the answer supposedly isn't 0.32 J or -0.32 J.

    The way you answered it makes perfect sense though...

    I think you need to incorporate the 7 cm, but I'm not sure how.
     
  8. Mar 1, 2009 #7

    LowlyPion

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    The limits of your integration are from .10 - .07 = .03 to .14 - 07 = .07.

    -1/2k*x2 = 1/2*200*(-.032 + .072)
     
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