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Work done to move spring displacement

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A spring has a relaxed length of 5 cm and a
    stiffness of 95 N/m. How much work must you
    do to change its length from 2 cm to 10 cm?

    k=95
    Lnull=0.05
    delta x = .1-.02 = 0.08


    2. Relevant equations

    F=-kx
    W=Fdcostheta


    3. The attempt at a solution

    I honestly have tried everything and am beginning to think I am way off the mark and missed something. But what i tried was

    W = Fdcostheta where F = -kx

    Therefore, since force changes direction after the displacement is past the relaxed spring length, i used:

    W = -95 * (0.05-0.02) * (0.05-0.02) cos 0 + -95 * (0.1-0.5) * (0.1-0.5) cos 180


    But that was apparently completely wrong. any help please?
     
  2. jcsd
  3. Oct 15, 2011 #2
    [itex] F = F d \cos \theta [/itex] is only valid if the force is constant over the distance (not a function of x in this case). Your force is a function of x, so you will have to integrate to get the work. It's possible you can solve the problem with a energy approach if integrals are beyond your course material.

    [tex] W = \int F(x) dx [/tex]
     
  4. Oct 15, 2011 #3
    EDIT: nvm realized my mistake was suppose to subtract

    Thanks a lot for reminding me force is not constant XD
     
    Last edited: Oct 15, 2011
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