Answer to Question #240836 in Physics for Neji

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

The work done by gravity:

$$W_g=mgh=mgL\sin\theta.$$

A fraction of this work is spent to overcome friction:

$$W_f=-\mu mg\sin\theta·L.$$

The work done by the man on the piano:

$$W_m=-FL=-fL$$

(because the force he exerts is equal to the force of friction),

$$W_m=-\mu mg\sin\theta ·L.$$

The net work:

$$W_\text{net}=W_g+W_f+W_m=\\ =mgL\sin\theta-2\mu mgL\sin\theta=\\ =mgL\sin\theta·(1-2\mu)=1093\text{ J}.$$