Work and force problem (piano moving on an incline)

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Homework Help Overview

The problem involves a 280-kg piano sliding down a 30-degree incline, with a man exerting a force to prevent acceleration. The context includes concepts of work, force, and friction, specifically addressing the effective coefficient of kinetic friction of 0.400.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to find the components of gravitational force along and normal to the incline, as well as the force of friction. Questions arise regarding the calculations for the force exerted by the man and the work done by various forces.

Discussion Status

Some participants have made attempts to calculate the forces involved and the work done, while others provide feedback on these calculations. There is an ongoing exploration of the signs in the calculations and the implications of the forces acting on the piano.

Contextual Notes

Participants are navigating the complexity of the problem, including the need to consider the direction of forces and the effects of friction. The discussion reflects uncertainty regarding the correct application of formulas and the interpretation of results.

balletgirl
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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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