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Help with spring stiffness calculation (k)

  1. Sep 18, 2016 #1

    I was given the following problem in my coursework:

    1. The problem statement, all variables and given/known data
    "A spring is initially compressed by 50mm when a steel ball of mass 2kg is released from just being in contact with the uncompressed spring. Determine the spring stiffness (k) of the spring."

    2. Relevant equations
    F = mg
    Restorative force F = -kx

    3. The attempt at a solution
    I gave the answer below, but the teacher has marked it as wrong. Can anyone shed any light on where I may have made a mistake? I thought it was just a simple application of Hooke's Law.

    F = mg
    F = 2 x -9.81
    F = -19.62 N

    F = -kx
    19.62 = -k(-0.05)
    19.62 = 0.05k
    k = 19.62/0.05
    k = 392.4

    Thanks, any insight gratefully received.
  2. jcsd
  3. Sep 18, 2016 #2


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  4. Sep 18, 2016 #3
    Thanks for your reply Bystander,

    I did actually include the units (N/m) in my coursework - sorry, I omitted them in my original post by accident.
  5. Sep 18, 2016 #4


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    I think the inclusion of "initially" implies the compression given is the lowest point the weight reaches rather than the equilibrium position that your solution assumes.
    Try applying conservation of energy to solve.
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