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

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?
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 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: $$m_1gh+0.5m_1v_1^2=m_2gh+0.5m_2v_2^2$$ That should help you out.
 But, since the height of m2 is set to zero, m2gh drops out. So then I'm left with my original formula, right?

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

 Quote by ph123 But, since the height of m2 is set to zero, m2gh drops out. So then I'm left with my original formula, right?
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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