Hi, this my first time asking a question, so here it goes: A 255 kg piano slides 4.6 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-35). The effective coefficient of kinetic friction is 0.40. (a) Calculate the force exerted by the man. So far, I have this answer, by subtracting the Fparallel(mg*sin(30)) by Ffriction(mu*mg*cos(30)). [383.8 N] (b) Calculate the work done by the man on the piano. Got this answer, too, by multiplying the answer above by 4.6m. [-1764.6 J] It's the next parts I got wrong. (c) Calculate the work done by the friction force. I figured Ffriction = mu * m * g * cos(30) = 865.7 * d = 3982.12 J, but that doesn't make sense. [?J] (d) What is the work done by the force of gravity? 9.8 m/s^2 * 255g * 4.6m = 11495.4, it's incorrect [?J] (e) What is the net work done on the piano? This one I had no idea how to do altogether. J Please help me out, thank you in advance!