Work and force problem (piano moving on an incline)

AI Thread Summary
A 280-kg piano is sliding down a 30-degree incline, and a man is pushing it back to prevent acceleration, with a coefficient of kinetic friction of 0.400. The discussion focuses on calculating the force exerted by the man, the work done by him, the work done by friction, the work done by gravity, and the net work on the piano. The user attempts to find the work done by each force, using the equation W = F*d, and calculates the force exerted by the man as approximately -421.5 N. The work done by the man is calculated as -1812.45 J, indicating that the direction of the force and movement are opposite. The calculations are being refined, particularly regarding the signs of the forces involved.
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Homework Statement



A 280-kg piano slides 4.3 m down a 30 degree incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is .400. Calculate:

(a) the force exerted by the man,
(b) the work done by the man on the piano,
(c) the work done by the friction force,
(d) the work done by the force of gravity, and
(e) the net work done on the piano.


Homework Equations



W=Fd... ?

The Attempt at a Solution



I know it's a long problem... I don't know how to find the work done by each force.
 
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You will need to find the components of the gravitational force that are along the ramp and normal to the ramp. Then you can find the force of friction.
The work done by any force is W = F*d.
 
a) Do I use mu*mg*cos(angle) - mg*sin(angle) ...?

If so, I did .40(280)(9.8)cos(30) - 280(9.8)sin(30) = -421.5 N
(Isn't it negative because it's going down?)


b) For this answer I did -421.5*4.3 and got -1812.45 J

I'm still trying to figure out the rest...
 
Looks okay, though I would have reversed the signs for the (a) calculation - the friction term reduces the force. But I agree with the minus sign in (b) since the piano is pushing on the man.
 
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