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Atwood with Sliding mass and real pulley

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    Block 1 with mass m1=4.04 kg rests on a very low friction horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to the hanging block 2 of mass m2=2.02 kg, as shown.

    The pulley is a uniform disk of radius 11.85 cm and mass 1.980 kg. Calculate the speed of block 2 after it is released from rest and falls a distance of 1.84 m.

    What is the angular speed of the pulley at the instant when block 2 has fallen a distance of 1.84 m ?

    2. Relevant equations
    Wtot=change in Energy
    KE=1/2 mv^2
    W=integral of force*displacement
    N2L for rotation and translation

    3. The attempt at a solution
    The tension of 2 and 1 on the pulley should be different but the accelerations of the blocks would be the same?
    a=m2*g/(mp/2 +m1+m2)
    T2=m2(g-a)=13.27
    T1=m1*a=13.0829
    WEm2=integral from 0 to 1.84 (T2xdx) = 22.47 J
    WT1m1=integral from 0 to 1.84 (T1xdx) = 22.1467 J
    work energy theorem:
    KE=W
    1/2mv^2=WEm2+WT1m1
    v=sqrt(2(WEm2+WT1m1)/mp)=3.33m/s

    This is not the correct answer, I'm not sure what I am doing wrong would the energy side of the equation be only kinetic?
     
  2. jcsd
  3. Apr 3, 2017 #2
    I think the diagram would be helpful. Is the pulley frictionless? Also why would the accelerations be same?
     
  4. Apr 3, 2017 #3
    m1 and m2 are atached by a string on a pulley so wouldn't their accelerations be the same? Screen Shot 2017-04-03 at 12.21.06 AM.png
     
  5. Apr 3, 2017 #4
    And all the information we were given is on the question so I'm assuming its not frictionless
     
  6. Apr 3, 2017 #5
    Correct. I can't really understand what the equations for work you've written, up you can just use conservation of energy to do the problem.
     
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