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Work, Energy, or Work problem

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    WorkEnergyPowerQuiz2picture.png
    The two masses in the Atwood’s machine shown in Figure 8–23 are initially at rest at the same height. After they are released, the large mass, m2 falls through a height h and hits the floor, and the small mass, m1rises through a height h. Find the speed of the masses just before m2 lands if h = 1.2m, m1 = 3.7kg and m2 = 4.1kg


    2. Relevant equations

    E=Eo
    Ug=mgh
    K=(1/2)mv^2


    3. The attempt at a solution

    m1*g*y +m2*g*y = (1/2)*m1*v^2 + (1/2)*m2*v^2

    I factored outed g*y and (1/2)*m*v^2
    g*y*(m1 + m2) = (1/2)*(v^2)*(m1 + m2)

    I cancelled out (m1 + m2)
    g*y = (1/2)*v^2

    g*y = (1/2)*v^2

    2*g*y = v^2

    √(2*g*y) = v

    √(2*9.8*1.2) = v

    This is apparently wrong:
    4.85m/s ≈ v
     
  2. jcsd
  3. Oct 12, 2011 #2
    first define the reference level for the potential energy..... what level have you chosen ?
     
  4. Oct 12, 2011 #3
    I would like to set the reference level at where the dotted line is (where v=0).
     
  5. Oct 12, 2011 #4
    ok, so that means your initial total potential energy is zero. Also initial total kinetic energy
    is zero. So total initial energy is zero. What about the final configuration ? The left block has climbed up distance h , so its potential energy is [itex]m_1gh[/itex] and the second
    block is gone below the reference level. so its potential energy would be
    [itex]-m_2gh[/itex]. What about their kinetic energies ? Find that expression and use the conservation of energy theorem, which implies that the total initial energy must be equal to the total final energy.......
     
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