The document contains calculations to solve for the potential energy (EPB) in three different systems. It first solves for EPB = 61 J. It then solves for the kinetic energy (ECB) = 41 J. Finally, it solves again for EPB = 52 J, finding the potential energy in three separate scenarios using the law of conservation of energy and given values for mass, velocity, gravity, and height.

1. EMA = EMB
ECA+EPA = ECB+EPB
0 + 140 = 79 + EPB
EPB = 61 J
EMA = EMB
ECA+EPA=ECB+EPB
29 + 18 = ECB + 6
47 -6 = ECB
ECB = 41J
EMA
ECA + EPA
(m.V^2)/2 + m.g.h
(2.8^2)/2 + 2.10.9
64 + 180
244 J
EMA = EMB
ECA+EPA=ECB+EPB
0 + 110 = 58 + EPB
EPB = 110-58
EPB = 52 J
Energ�a Mecanica II
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