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Calculate the speed of the Cylinder in the pulley system

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Each of the two systems is released from rest. Calculate the speed v of each 47-lb cylinder after the 38-lb cylinder has dropped 5.2 ft. The 17-lb cylinder of case (a) is replaced by a 17-lb force in case (b).

    I have attached an image of the question.

    2. Relevant equations



    3. The attempt at a solution

    I think I should use U1 + K1 = U2 + K2

    But I'm not quite sure how I should interpret the situation with this formula. I had thought:

    47lbf*5.2ft - 55lbf*5.2ft = 0.5*m*v2

    But I'm not sure if this is the right way to look at it.

    Any help is appreciated
     

    Attached Files:

  2. jcsd
  3. Feb 24, 2013 #2
    I've managed to figure out part A

    m1gh = m2gh + .5m1v2+.5m2v2

    (47lb)(5.2ft) = (55lb)(5.2ft) + .5(47lb/32.2)v2 + .5(55/32.2)v2

    solving for v gives me:

    v = 5.1187 ft/sec

    I had thought that for part B I would be able to use the same equation but use a reduced mass, 38 instead of 55 but it doesn't work.

    Input on this would be greatly appreciated.
     
  4. Feb 24, 2013 #3

    haruspex

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    Something crossed over there - that would give a negative value for the gain in KE. I guess that was just an error in the post.
    That should work. Pls post the details.
     
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