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Hookes Law

  1. Oct 1, 2011 #1
    Okay so we have a block attached to a spring with a spring constant of 50 N/m. The spring is in the relaxed state.
    We apply a force 3N along the positive x direction. The block moves along the x axis,stretching the spring until the block finally stops.
    What is the block's displacement?

    When the block is stopped the force of the spring on the block should be -3 Newtons. Using the formula F=-kx
    I get the answer 3/50=0.06m. The book's answer is 0.12m.
    What exactly am I doing wrong?
     
    Last edited: Oct 1, 2011
  2. jcsd
  3. Oct 1, 2011 #2

    SammyS

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    I'm not exactly sure since you haven't shown your work.

    However I suspect the problem has to do with the fact that when the block gets to the position at which the applied force is cancelled by the force exerted by the spring, at this position the block is moving and continues to move due to its inertia.
     
  4. Oct 1, 2011 #3

    PeterO

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    Your force will accelerate the block, until it has stretched the spring 6 cm. At that time the mass will be moving, and will take some time to stop.

    Imagine the x direction was down, and you had a mass of weight 3N attached to the spring then dropped. It will continue past the point where the spring pulls with 3N. It will eventually stop and rebound and you are basically asked where will it eventually stop.
     
  5. Oct 1, 2011 #4
    Oh, okay I understand now thanks!

    I actually found a way right before checking this post, using the energy equations.
    W = Fx
    E = -.5kx^2
    Fx = -.5kx^2
    F = .5kx
    x = -2F/k
    x = -2*-3N/50 = .12m

    So if I want the final state of a spring but I don't want to do calculations involving inertia and oscillation, I should just use energy principles then?
     
  6. Oct 1, 2011 #5

    PeterO

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    I thought of giving a work done vs Energy stored in the spring option but was not sure if you had done that topic. Clearly you had, and that is the easiest way to solve.
     
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