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Dropping a ball on a spring.

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A massless pan hangs from a spring that is suspended from the ceiling.
    when empty the pan is 50cm below the ceiling. If a 100g clay ball is placed gently
    on the pan, the pan hangs 60cm below the ceiling. Suppose the clay ball
    is dropped from the ceiling onto an empty pan. What is the pans distance
    from the ceiling when the spring reaches its maximum point.
    3. The attempt at a solution
    So first I find the spring constant
    [itex] mg=kx [/itex]
    k=9.8 N/m
    Now we find how much the spring stretches when the clay ball drops on it.
    [itex] mgh=\frac{kx^2}{2} [/itex]
    h=.5m
    when I solve for x i get x=.31m
    so we add this to .5m and we get 81cm
    the answer in my book is 93cm which is basically off by 10cm from my answer
    Do I need to add 10cm to mine to account for the weight of the clay ball.
    but it seems that is taken into account with the potential energy.
     
  2. jcsd
  3. Jul 21, 2012 #2

    ehild

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    The ball falls x more than 50 cm.

    ehild
     
  4. Jul 21, 2012 #3
    ok thanks for your post so
    it needs to be [itex] mg(h+x)=\frac{kx^2}{2} [/itex]
     
  5. Jul 21, 2012 #4

    ehild

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    Right. :smile:

    ehild
     
  6. Jul 21, 2012 #5
    its already has an extension of 50 cm so it wud be mg(h +x)= kx^2/2
     
  7. Jul 21, 2012 #6
    Can i reason it this way as calculated by OP?

    If i assume it to be on frictionless surface and given a kinetic energy of mgh.
    The spring will extend to an equation of mgh=1/2(kx2)
    The weight will produce additional 10cm(given)
    Thus total height is 31+10+50.

    Any flaws in my thinking.
    Thank you.
     
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