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Energy Loss Question

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A (smooth) rope of length L and mass m is placed above a hole in a table. One end of the rope falls through the hole, pulling steadily on the remainder of the rope. Find the velocity of the rope as a function of the distance to the end of the rope, x. Ignore friction of the rope as it unwinds. Then find the acceleration of the falling rope and the mechanical energy lost from the rope as the end of the rope leaves the table. Note that the rope length is less than the height of the table.


    2. Relevant equations



    3. The attempt at a solution
    Well I already got the right equations for v and a....

    v(x) = (2gx/3)^.5
    a(x) = [(2g/3)^.5] / 2√x

    But I don't understand how to find the mechanical energy loss. It seems like mechanical energy shouldn't be lost here, because the lost PE is just changing into KE...
     
  2. jcsd
  3. Mar 18, 2014 #2

    haruspex

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    This is a tricky question that comes up regularly in various guises.
    It should state that the rope is inelastic. This means something interesting happens as each new piece is accelerated from rest, apparently instantaneously, to the current speed of the descending part. This constitutes an impulse, so momentum is conserved but not work.
     
  4. Mar 19, 2014 #3
    So the change in mechanical energy would be the difference between the sum of all kinetic energies of small sections of the rope and the change in Potential Energy? But how would you know the change in PE if they don't give you a table height?

    I got mgL/6 for the sum of all the KE of all small sections of the rope.
     
  5. Mar 19, 2014 #4

    haruspex

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    Once a section of rope has passed through the hole there will b no further loss of work, so the height does not matter.
    Let the length of the rope that has passed through the hole be x and the current speed of that section be v. Consider a small section length dx passing through the hole, going from rest to speed v, and apply conservation of momentum.
     
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