Question

A 330 kg piano slides 3.6 m down of 28° incline. The piano is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40. Calculate the net work done on the piano. (The piano is kept from accelerating means that the piano is NOT accelerating. It is moving at a constant velocity).

Answer 1

$W_{friction}=F_{friction}\cdot d\cdot \cos180°=\mu mg\cos28°\cdot d\cdot \cos180°=$

$=0.4\cdot 330\cdot 9.8\cdot\cos28°\cdot3.6\cdot \cos180°=-4112\ (J)$

$W_{p}=F_{p}\cdot d\cdot \cos180°=(mg\sin28°-\mu mg\cos28°) \cdot d\cdot \cos180°=$

$=(330\cdot9.8\cdot\sin28°-0.4\cdot330\cdot9.8\cdot\cos28°) \cdot 3.6\cdot \cos180°=-1354\ (J)$

$W_G=mg\cdot d\cdot\sin28°=330\cdot9.8\cdot3.6\cdot\sin28°=5466\ (J)$

$W_{total}=5466-1354-4112=0\ (J)$