Homework Help: 2 weights on a pully

1. A 3kg weight is attached to a 2kg weight by a rope. This rope is placed over a pully so the weights are hanging. The 2kg weight starts 4 meters lower than the 3kg weight. If the system starts at rest what is the speed of the objects when they are at the same height.



2. Fnet = ma
v^2=2*a*(x2-x1)




3. I took the net force to be the normal force of the 3kg weight minus the normal force of the 2 kg weight. Fnet=9.81N I then divided by 3kg to get a=3.27 m/s^2 I then plugged knowns into v^2=2*a*(x2-x1) v^2=2*3.27*2 and got v=3.617 This answer does not work. What am i doing wrong?