Finding the angle of an inclined plane

[scimanyd]

This should be an easy question, but I am having problems with my conflicting solutions. Maybe someone could help.

A 1900 kg car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down the hill at 27 m/s. Going up a hill, the car's engine needs to produce 47 hp more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

My approach:

First P=Fv ... so I sub in Force =Power/velocity
and then set up my force equations in the x direction which I set parallel to the inclined plane.

Assuming Power up = power down + 47hp(or 35060Watts)

UP: P/v - sinθmg - (friction+air resistance) = 0
DOWN: (P+47hp)/v + sinθmg - (friction+air resistance) = 0

I tried setting these two equations equal to each other and I come up with theta <1°
I tried solving for P in one and then subbing into the other and solving for Theta and I get theta = about 2°.
Still this seams unreasonably low, I wouldn't think that it would take 47hp more to go up a 2° incline

 

[NateTG]

Can you specify what the variable 'P' is supposed to represent? Also, are you sure that the first term in both of your expressions makes sense?

 

[scimanyd]

Yes, P stands for power
I substituted P/v in for my applied force, since power = Force*velocity
I should be allowed to make the sub.

the equations simplified: Force - weight(portion along x axis) - Friction = 0 ... (constant velocity tells me that the equation is in equalibrium)

 

[neo143]

In second equation instead of
P+47 you had to write P-47

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