net work done on the piano

Question

A 330 kg piano slides 3.6 m down of 28o 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

Answer 1

W_friction​ = F_friction​ ⋅d⋅ cos180° = μmg cos28° ⋅d⋅ cos180°

= 0.4 ⋅ 330 ⋅ 9.8 ⋅ cos28°⋅ 3.6 ⋅ cos180° = −4112 (J)

W_p​ = F_p ​⋅d⋅ cos180° = (mg sin28° − μmg cos28°) ⋅d⋅ cos 180°

= (330 ⋅ 9.8 ⋅ sin28° − 0.4 ⋅ 330 ⋅ 9.8 ⋅ cos28°) ⋅ 3.6 ⋅ cos180° = −1354 (J)

W_G​ = mg ⋅d⋅ sin28° = 330 ⋅ 9.8 ⋅ 3.6 ⋅ sin28° = 5466 (J)

W_total​ = 5466 − 1354 − 4112 = 0 (J)