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Energy conversation and friction

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data
    At a construction site, a 65 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 80.0 kg box on a horizontal roof. The cable pulls horizontally on the box, and a 50.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown.

    A. Find the friction force on the bag of gravel and on the box.
    B. Suddenly a worker picks up the bag of gravel. Use the energy conservation to find the speed of the bucket after it has descended 2.0 m from rest.

    http://img237.imageshack.us/img237/8178/yf0729.jpg [Broken]

    2. Relevant equations

    Ki + Ui + Wdissipative = Kf + Uf

    3. The attempt at a solution

    The frictional force on the gravel is 0 because it's not on the ground. For the box it's (130 kg)(9.8 m/s2)(0.7) = 891.8N.

    Initial kinetic energy: 0
    Initial gravitation potential energy: (65 kg)(9.8 m/s2)(2.0 m) = 1274 J
    Dissipative work: (80.0 kg)(9.8 m/s2)(0.4)(2.0 m) = 627.7 J
    Final kinetic energy: ?
    Final gravitational potential energy: 0

    Ki + Ui + Wdissipative = Kf + Uf
    0 + 1274J + 627.7J = Kf + 0
    Kf = 1901.7 J
    1901 J = (145 kg*v2)/2 ==> v = 5.12 m/s

    Where did I go wrong? By the kinematics equations, it should be 2.99 m/s.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 24, 2010 #2


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    The two coefficents of friction are for standing and kinetic I hope.

    only mistake i see is the "Ki + Ui + Wdissipative = Kf + Uf" If it is disspated it is on the right side.

  4. Feb 24, 2010 #3
    Yes, that's it. I guess I copied that equation down incorrect. Thank you very much. :smile:

    Here are the corrected calculations for anyone else with this problem:

    Ki + Ui = Kf + Uf + Wdissipative
    0 + 1274J = Kf + 0 + 627.7J
    Kf = 646.3J
    646.3J = (145 kg*v2)/2 ==> v = 2.99 m/s
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