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Springs Equilibrium Length Question

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    The equilibrium length of a certain spring with a force constant of K = 260N/m is 0.25m

    What is the magnitude of the force that is required to hold this spring at twice its equilibrium length?

    2. Relevant equations

    Hookes Law F=Kx

    3. The attempt at a solution

    260 N/m = F/.25
    twice the length is .50m so 260 N/m * .50 = 130
  2. jcsd
  3. Nov 23, 2008 #2
    Here's a picture:


    The equilibirium, original, X=0 point in this case is where X=0.25m. You're looking to extend the spring to 0.5m in length - that's an extension of 0.25m. So I think the correct figure should be 260 * 0.25 = 65N
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