A330-kg piano slides 3.6 m down a 28 \incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. the coefficient of friction is 0.40. calculate the force exerted by the man

he role of "man" is to just stop it from getting accelerated let F be the force by man up along >. net force on block = m a (down) = mg sin 28 - F - uk(mg cos 28) since acceleration a = 0 mg sin 28 = F + uk(mg cos 28) F = mg[sin28 - 0.4 cos 28] = 376.1 N work done by man = F * d cos 180 = - F *3.6 = - 1353.96 J -ve sign because F and displacement are 180 deg apart ... work done by friction = - 0.4 {mg cos 28} * 3.6 = - 4111.85 J ... work done by gravity = + {mg sin 28} * 3.6 = 5465.78 J net work done = net force *3.6 = 0 *3.6 = 0 there is no net force on piano net work done = 5465.78 - 1353.96 - 4111.85 = 0 (very very close to zero) because decimal places??