Stopping 15000 kg Truck with Friction and Gravity

[marissa12]

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 deg. and the coefficient of friction is 0.40.

Use work and energy to find the length of a ramp that will stop a 15,000 kg truck that enters the ramp at 35m/s

Delta(kinetic energy)=W(net)

W(friction)= -u_k*m*g*d(distance)
W(gravity)= -m*g*(delta y)

0.5*m*v^2= W(friction) and then you solve for d.. is this right? because i got some really weird answers

 

[zwtipp05]

Find y in terms of d using trig and then do W(Friction) + W(gravity) = KE

 

[marissa12]

maybe i screwed up the math..because that didnt work

0.5*m*v^2=(-u_k*m*g*D)-(m*g*D*sin(6))
9187500=-58860d-146344d
9187500=-205204d
d=12.4

and that was still wrong, and i can't see where i messed up in the math