A 330-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

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W0lf93
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??