Runaway Truck: Energy and Work

[LinearBeat]

Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of friction in the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0° and the coefficient of friction is 0.46. Use work and energy to find the length of a ramp that will stop a 10,000-kg truck that enters the ramp at a speed of 35 m/s (75 mph).

F_f=Force of Friction (F_f=μF_n; F_n=Normal Force)
m=Mass of the truck
U=Potential Energy, U-i= initial U, U_f=final U
L=Length of the ramp
K=Kinetic Energy, K_i=initial K, K_f=final K
W=Work (in this case done by friction)
v=initial velocity of the truck

Looking at the diagram I drew I determined:
F_f=μmg/cosθ
U=Lsinθmg

ΔK=ΔU-W
K_f-K_i=U_f-U_i-W
-K_i=U_f-W
-1/2mv^2=Lsinθmg-(μmg/cosθ)L
-1/2v^2=L(sinθg-μg/cosθ)
-v^2/(2(sinθg-μg/cosθ))=L

Using this process I've gotten the wrong answer, but I don't understand what I've done wrong and how I need to revise it.
Thanks!

 

[Villyer]

I think the problem may lay in your equation where you related energies.

Your statement -K_i=U_f-W is saying that all of the initial kinetic energy is being transformed into work, except for the potential energy that is being gained (dividing through by a negative gives K_i=W-U_f). This is also saying that initial kinetic plus initial gravitational results in the work done by friction, which seems incorrect.

 

[LinearBeat]

Okay so that's a great way to evaluate my original statement. The way I understand it, the kinetic energy that the truck has will all be converted into potential energy and work. So I think a better equation is:
K_i=U_f+W
Using Villyer's method I think this equation holds up; the kinetic energy minus the work should equal the potential energy and the kinetic energy minus the potential energy should equal the work.
and the verdict is...
Correct!
Thanks so much Villyer.