Calculating Net Work on a Sliding Piano

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A 260 kg piano slides down a 30° incline, with a man exerting a force to prevent acceleration. The force exerted by the man is calculated to be approximately 391.35 N, while the work done by him is about -1682.79 J, and the work done by friction is -3795.41 J. The discussion highlights the need to calculate the work done by gravity and the net work on the piano, which remain unsolved. Participants emphasize understanding the forces at play, particularly how the net force relates to the system's equilibrium. The conversation focuses on applying the work formula and analyzing the forces involved to find the missing values.
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Homework Statement



A 260 kg piano slides 4.3 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40.

Figure 6-36

(a) Calculate the force exerted by the man.
391.3469084 N
(b) Calculate the work done by the man on the piano.
-1682.791706 J
(c) Calculate the work done by the friction force.
-3795.408294 J
(d) What is the work done by the force of gravity?
? J
(e) What is the net work done on the piano?
? J



Homework Equations


Work=Force*Distance
gravity=9.8
Fnet=Sum of all forces

The Attempt at a Solution


Letters a through c are correct, but I have not been able to solve for d and e. I have tried doing W=FX for letter d, but with no luck. I cannot solve for e without solving for d.

Any help is appreciated. Thank you!
 
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Kinetic friction cannot do any work.
 
Think of the formula

Work = force * distance

Now it's kept from accelerating, what does that mean about your net force?

Once you have that you can assume that Fnet= Fpiano + FFriction + Fperson.

So all you need to find to solve this problem is the horizontal force applied by the piano. After that you just plug stuff in. So think about what is making the piano move down, then what component of that the person would be resisting.
 
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