The coefficient of friction (uk) between the 3.00 kg block (m1) and the surface in Figure P7.21 is 0.200. The system starts from rest. What is the speed of the 5.00 kg ball (m2) when it has fallen 1.20 m (y)? Wnet = [tex]\Delta[/tex]KE Um2 - Wfk = .5m2vf2 - .5m2vi2 (m2g)(y) - (uk)(N)(x) = 5m2vf2 - .5m2vi2 (m2g)(y) - (uk)(m1g)(x) = .5m2vf2 - .5m2vi2 m1 = 3 m2 = 5 uk = .2 y = 1.2 vi = 0 vf = ? x = ? I'm trying to find vf. I don't know what x would be for the work done by friction. I'm sure I made some other mistakes as well. Thanks in advance.