Kinetic Friction and skis thrust

[BlakeGriffin]

Homework Statement

Sam, whose mass is 71.0 kg , takes off down a 51.0 m high, 11.0 degree slope on his jet-powered skis. The skis have a thrust of 210 N . Sam's speed at the bottom is 41.0 m/s .

Homework Equations

I'm not sure which equation(s) to use.
Net Force= ma and some kinematic equations maybe?

The Attempt at a Solution

This is the free body diagram that I came up with.
http://img291.imageshack.us/i/fbdy.jpg/ (It's not to scale)
There is also the thrust force coming from the skis in the opposite direction of the friction force. I couldn't fit them all in. Btw the dashed lines are the components of the gravity force.

If my free body diagram is right, then normal force = FG Cos(theta) which is one of the components of the gravity force. The other component of gravity is FG Sin(theta)

FG is mg = (71 KG)(9.8 m/s^2)=695.8 N
FN=FG Cos(theta)=695.8(cos(11))0=683N
The other component of FG is =FGSin(theta)=695.8(Sin(11))=133N

Since this component of gravity and the thrust force coming from the skis both point down the hill (or whatever) I get 133+210=343N pointing in that direction.

I don't know where to go from here.. Can anyone help me?

 

[amy andrews]

Ummm...what's the problem asking? Is it looking for the coefficient of friction or something?