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Work done on a spring

  1. Feb 24, 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 150 N/m. How much work must you do to change its length from 12 cm to 15 cm?

    2. Relevant equations

    F=-kx, W=F(delta)x

    3. The attempt at a solution

    ?? ((.12 m)-(.07 m) * (150 N/m)) = 7.5 N

    ((.15 m)-(.07 m) * (150 N/m)) = 12 N

    (12-7.5 N) * (.15-.12 m) = .135 Nm
     
  2. jcsd
  3. Feb 24, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    You need to consider that W = ΔPE

    What is the formula for the PE of a spring?
     
  4. Feb 24, 2009 #3
    thanks much for the help!
     
  5. Feb 24, 2009 #4
    i assume you got the answer but if not
    a quick way to solve this problem is to
    set up a integral.
    integrate Kx where K is the stiffness
    so it gives you 75x^2
    act like relaxed lenght is initial (or 0) then find difference
    from initial your two asking points are (make sure your in m not cm)
    ex. rest is 7cm find work 13cm stretched to 15 cm
    first limit would be (.13-.07)=.06 next (.15-.07)=.08
    just plug these in to 75x^2
    do F(b)-F(a) and that gives you the answer
     
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