Question regarding distance traveled up an incline?

Homework Statement

After a bobsled race, the sled and riders have been partially slowed down up in an icy incline, but they need to be brought to a stop. This happens on a portion of track inclined at 30 degrees that has a coefficient of kinetic friction of 0.60. The sled enters the incline at 25.0 m/s. How far does it travel along the incline before stopping?

Homework Equations

(1/2)mv(final)^2 - (1/2)mv(initial)^2 = Ffriction

The Attempt at a Solution

Im not sure how to incorporate the angle of the incline, but as an attempt, i did
1/2vi^2=FG(cos 30) d
and i got 88.3 m. I'm not sure if that's right...

Doc Al
Mentor
Looks like you're using an energy approach. Good. But don't forget gravitational PE, which increases as the sled goes up the incline.

How would you incorporate gravitational PE? and also, would the formula
d = vi^2 / [2g(sin30 + ucos30)] work?

Doc Al
Mentor
How would you incorporate gravitational PE?
What's the definition of gravitional PE? How do you calculate it?
and also, would the formula
d = vi^2 / [2g(sin30 + ucos30)] work?
I would advise against looking for a formula to plug into unless you fully understand what the formula means. Better to stick with the basics and derive your own formula, tailor-made to the specific problem. (That formula is close, but not quite right, anyway.)

The basic idea that you want to use is energy conservation:
Energy(initial) + Work done by friction(which is negative) = Energy(final)

Here "Energy" means total mechanical energy: KE + PE.