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Finding the masses of two blocks in a pulley system using work-energy theorem

  1. Mar 14, 2007 #1
    Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.00m , its speed is 1.50m/s . If the total mass of the two blocks is 13kg, then what is:
    1)the mass of the more massive block
    2)the mass of the less massive block

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    Since the more massive block descended 1m, then the less massive block must have ascended 1.0m. Also, since the more massive block travels at 1.50m/s, then the less massive block must trave -1.50m/s. Setting the height of the less massive block, m1, as 2.0m, and the height of the more massive block, m2, as 0, then

    m1gh + (1/2)m1v1^2 = (1/2)m2v2^2
    m1(gh + (1/2)v1^2) = m2(1/2 *v2^2)
    m1/m2 = v2^2/(gh + v1^2)
    m1/m2 = (1.125 m^2/s^2) / (20.725 m^2/s^2)
    m1/m2 = 0.0543

    I figured, since given only the combined mass, I had to find the ratio of the masses. I multiplied the ratio (0.0543) times 13kg, and found that m1 = 0.706kg, and 13kg - 0.706kg = 12.294kg. Not right, though. Anyone know where I went wrong?
     
  2. jcsd
  3. Mar 15, 2007 #2
    Yeah. Your equation for conservation of energy isnt right. You didnt factor in the loss of gravitational potential energy of the heavier mass. It should be:

    [tex]m_1gh+0.5m_1v_1^2=m_2gh+0.5m_2v_2^2[/tex]

    That should help you out.
     
  4. Mar 15, 2007 #3
    But, since the height of m2 is set to zero, m2gh drops out. So then I'm left with my original formula, right?
     
    Last edited: Mar 15, 2007
  5. Mar 15, 2007 #4

    PhanthomJay

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    You are setting the final mechanical energy of m1 equal to the final mechanical energy of m2. This is not a correct application of the law of conservation of energy, which states that, in the absence of non-conservative forces doing work, that the initial mechanical energy of the system is equal to the final mechanical energy of the system. I would start by taking the initial potential and kinetic energy of the system to be zero prior to the release of the blocks, at h=0, and go from there.
     
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