Work and energy: truck and ramp?

Question

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 5.0° and the coefficient of friction is 0.48. Use work and energy to find the length of a ramp that will stop a 17,000 kg truck that enters the ramp at 35 m/s (75 mph).

I tried using the 1/2mv^2 + f(dx) = 1/2m(vcos(theta))^2 + mgh and I seem to be getting about 717 meters, which is wrong. I think I am finding dx wrong. How do you calculate dx? Is the equation right in the first place?

Thanks!

Answer 1

I would use conservation of energy.
x is the distance traveled on the ramp
the forces at work are:
Gravity. The potential energy will be
sin(5)*m*g*x

Friction:
.48*cos(5)*m*g*x

Since there are no brakes all of the kinetic energy will be lost to friction and the potential energy gain:

.5*m*v^2=
.48*cos(5)*m*g*x+sin(5)*m*g*x
note that m divides out
I get 110 m

j