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Ideal Spring

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    An ideal spring is hung vertically from the ceiling. When a 2.0 kg mass hangs from it at rest, the spring is extended 0.06 meters from its relaxed state. An upward external force is then applied to the block to move it upward a distance of 0.16 meters. While the block is being raised by the force, the work done by the spring is:

    2. Relevant equations
    Force from a spring= -kd
    Work due to a spring= .5(k)(xi)^2-.5(k)(xf)^2

    3. The attempt at a solution
    I believe to find the magnitude of the spring constant i rearrange F= -kd to become k=F/d (i read that we can neglect the negative in this case? correct me if im wrong)

    so F=mg=(2.0kg)(9.8m/s^2)=19.6N
    19.6N/.06m=326.667N/m=k

    Finally, W=.5(326.667N/m)(.06m)^2 - .5(326.667N/m)(.10m)^2= -1.04J

    Did i find the spring constant correctly? I feel like using mg/d was wrong.
    Also, when i use the work done by spring equation, is xi and xf the distance away from equilibrium?
     
  2. jcsd
  3. Apr 7, 2016 #2
    the k value is correct.
    regarding work done by the spring -pl. check whether the choice of xi and xf is correct....
    in the sector of extended portion the external force was supporting the wt. so what work was done by the spring -it came back to the relaxed state...i am just trying to think ....alongwith you
     
  4. Apr 8, 2016 #3

    haruspex

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    Yes, that's ok. In fact, the force is always opposite to the displacement, so technically it should be k=-F/d.

    That all looks correct. Are you surprised by the minus sign?
     
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