2 masses around a pulley w/ friction

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SUMMARY

The discussion focuses on calculating the final speed of mass m (8.69 kg) in a pulley system involving two masses (M = 15.3 kg) and a uniform disk pulley (Mp = 14.1 kg, radius R = 7.50 cm). The user initially attempted to apply conservation of energy but arrived at an incorrect speed of 3.67 m/s instead of the correct answer of 3.23 m/s. The correct acceleration formula derived is a = (g(M-m))/(M+m+(0.5*Mp)), which leads to the accurate calculation of the final speed after moving 2.50 m.

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Homework Statement



5.Two objects, M = 15.3 kg and m = 8.69 kg are connected with an ideal string and suspended by a pulley (which rotates with no friction) in the shape of a uniform disk with radius R = 7.50 cm and mass Mp = 14.1 kg. The string causes the pulley to rotate without slipping. If the masses are started from rest and allowed to move 2.50 m what is the final speed (m/s) of mass m?

Homework Equations



M= 15.3 kg
m= 8.69 kg
r= 0.075 m
Mp= 14.1 kg
d= 2.5 m
g= 9.8 m/s^2
i= 0.5*Mp*r^2 = 0.03965625




The Attempt at a Solution



I'm not sure what to do with the pulley... But my first try at finding the V of M was to use conservation of energy, which came out to be:

v=SQRT(2gh ((M-m)/(M+m)) ) = 3.674376423

The correct answer should be 3.23. What am I missing?
 
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Ok, I was able to get the acceloration using another thread from PF, and then from there, I calculated SQRT(2ah) and that gave me the correct velocity!

a = (g(M-m))/(M+m+(0.5*Mp))
 

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