What is the Coefficient of Kinetic Friction for an Inclined Object?

[RingWraith2086]

OK here's the problem: An extremely light drivable car with a mass of only 9.50 kg was built. Suppose that the wheels of the car are locked, so that the car no longer rolls. If the car is pushed up a 30° slope by an applied for of 80.0 N, the net accel. of the car is 1.64 m/s2. What is the coefficient of kinetic friction between the car and the incline?

And where's what I've done:
m = 9.5
W = 93.195
a = 1.64
Fnet = ma = 15.58
Fapplied = 80
Ffriction = Fapplied - Fnet = 64.46
Fperendicular = CosTheta(W) = 80.709 (also Fnormal)
Fparallel = SinTheta(W) = 46.5975

Then I have:
Coeff of friction = Ffriction/Fnormal = 64.46/46.5975 = 1.3833

But according to the answers my teacher gave me, and logic (that seems a little bit high for the situation), that isn't right. Please show me where I'm going wrong. Thanks.

 

[jamesrc]

You forgot the component of the weight along the incline when calculating F_net.

$$F_{\rm net} = F_{\rm applied} - F_{\rm parallel} - F_{\rm friction}$$

Try redoing it from there and see how it works out.

 

[RingWraith2086]

This teacher is the best science teacher I've ever had, but for some reason she failed to mention that the Fparallel was in the Fnet formula on the angle. And I know I didn't just copy it down wrong, because two of my friends (one of whom made a 36 on the math part of his ACT) couldn't do it either. I'll ask her about it when I get back to school, but thanks for the help. I've got another one I need help on, but I'm going to start a different thread. Thanks again.