Calculating Coefficient of Friction on Incline Pulled at Constant Velocity

[BunsenBurner1]

A sled is pulled 3 meters at constant velocity up a 20 degree incline. the efficiency of this procedure is 42%. Calculate the coefficient of friction between the crate and the incline.

alright I'm completely clueless on this one.
i know that efficiency=output (energy delivered by machine)/input (energy supplied to machine)
but i don't know what other equations I'm supposed put into play here as well.

 

[Shooting Star]

First think what are the forces resisting the motion upward. Then find their magnitudes.

 

[BunsenBurner1]

so I've got the force of friction, normal force, pulling force, and mg
Ff, Fn, Fp, mg

and KE and PE are in play here...but...KE can be ruled out and PE would be mg(1.026) so Fp=mg(1.026)?

 

[BunsenBurner1]

well i got .73 for mu but I'm not sure if that's right...

 

[Shooting Star]

If you show the calculation, I can verify it for you. It can't be too long.

 

[BunsenBurner1]

ok looking it over again i got a different answer mu = .86
Equation: Efficiency=Wout/Win where W=Force*distance

efficiency being 42%

distance i found by multiply 3*sin20 which led me to find Wo=mg(1.026)

to find Wi --> F=Nu (F being force, N being normal force, u being coefficient of friction)
N=mg(3*cos20)u
Wi= 2.82mgu

so i plugged them in: .42 = mg*(1.026) / mg*u*(2.82)
mg cancels out and I'm left with u= .86
right??

 

[Shooting Star]

to find Wi --> F=Nu (F being force, N being normal force, u being coefficient of friction)

Why is F only the friction here?

Energy input = increase in PE_grav + work done against friction.
Energy output = increase in PE_grav.

 

[BunsenBurner1]

Why is F only the friction here?

Energy input = increase in PE_grav + work done against friction.
Energy output = increase in PE_grav.

wouldnt the PE_grav. end up canceling each other out though?

 

[Shooting Star]

In x/(x+y), do the x's cancel each other?