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Kinetic Energy in a pulley system

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    http://s2.webstarts.com/exploring-my-thought/index.html [Broken]

    Find the kinetic energy of the system of bodies B1, B2 and B3 at an instant when the speed of B1 is 5ft/sec (the image is linked above)

    radius of B2 = 2ft
    mass of B1 = 32.2lb
    mass of B2 = 32.2lb
    mass of B3 = 96.6lb

    answer is 25ft-lb

    2. Relevant equations
    translational kinetic energy: K = 1/2mv2
    rotational kinetic energy: K = 1/2Iw2
    3. The attempt at a solution

    B1 = 1/2(32.2)(5)^2 = 402.5 ft-lb
    B2 = 1/2((mr^2)/2)(5/2) = 201.25 ft-lb
    B3 = 1/2(96.6)(5)^2 = 1207.5 ft-lb

    My answers are way off, what am I doing wrong?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 6, 2009 #2

    Matterwave

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    Science Advisor
    Gold Member

    How can energy be measured in ft-lb? That doesn't make sense unit wise.

    A ft-lb is a unit of torque...
     
  4. Dec 7, 2009 #3

    Work = force * distance right?

    force unit:lb
    distance unit :ft
     
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