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Work done with free body released

  1. Nov 5, 2007 #1
    I have a question here

    My attempts is
    mgx=(1/2)*kx^2 + (1/2)mv^2
    where mgx is the potential energy of block m
    and (1/2)mv^2 is kinetic energy of block 2m.
    But this does not seem to work, because we don't have the value v.

    May I know how to solve it?
    Last edited: Nov 5, 2007
  2. jcsd
  3. Nov 5, 2007 #2


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    Homework Helper

    you have the right equation... v is 0 when the object reaches maximum extension... because at this point, the hanging M mass will change from going downward to going upward.
  4. Nov 5, 2007 #3


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    Draw the free body diagram for M and 2M. From that find the common accelaration. Now work done the masses = potential energy stored in the spring. i.e. 3Max = 1/2*kx^2
  5. Nov 5, 2007 #4
    Thanks, rl.bhat, but I still getting the correct answer(22cm) if sub v=0 in
    mgx=(1/2)*kx^2 + (1/2)mv^2
    as mentioned by learningphysics.
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