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Hooke's Law spring stiffness (k) problem - help! Watch

    • Thread Starter

    Hi there

    I was given the following problem in my coursework:

    "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."

    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
    Restorative force F = -kx

    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

    Is this correct? Thanks.

    I think you've answered the question as if the ball were being gently lowered onto the spring until it reached the equilibrium point... which is a type of question which gets asked.

    This question seems to be more dynamic - if the ball is suddenly released it'll accelerate downwards and gain momentum... and travel past the equilibrium position. In an ideal system with no losses it'll be in SHM and will be stationary again for an instant when it's on the opposite side of the equilibrium position to where it started, that position is I think 50mm below the release position.
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