In the figure the second block slides horizontally without friction while the first block falls. Use the Conservation of Energy to find the speed of the blocks after they have moved 2 meters, starting from rest. The falling block has mass 3 kg, the sliding block has mass 5 kg, and the pulley is ideal.
See Figure 1.
Uf + Kf = Ui + Ki
The Attempt at a Solution
So for the 3 kg block:
mghf = 1/2mvf2 = mghi + 1/2mvi2
(3 kg)(9.8 m/s2)(2 m) + 1/2(3 kg)(vf)2 = 0
1/2(3 kg)(vf)2 = 58.8 J
(vf)2 = 39.2 m2/s2
vf = 6.26 m/s
I'm not sure if I quite got the point of the conservation of energy bit. Does it mean the work done by the pulley is zero? That's what I had to assume to get vf for the 3 kg block. I'm not sure what to do about the 5 kg block, though, because if I just use 2 m for the h for it, then I get the same vf as the 3 kg block, which seems odd to me. But maybe I'm missing the point.
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