Question

A truck with failed brakes is moving downgrade at 120 km/hr just before the driver steers the truck up an emergency escape ramp with an inclination of theta= 15 degrees. The trucks mass is 1.4x10^4 kg. Assume that the rolling friction behaves like kinetic friction, and the coefficient of rolling friction between the tires and ramp is 0.10. What minimum length L must the ramp have if the truck is to stop (momentarily) along it?

Answer 1

u =120 km/hr = 33.33 m/s

m = 1.4*10^4 kg, uk =0.1

Fk = uk*N = uk*mgcos(theta)

sin(theta) = h/L

Work done by friciton force

W = 0.5mu^2 -mgh

m*uk*gL*cos(theta) = 0.5mu^2 - mg*L*sin(Theta)

0.1*9.8*L*cos(15) = 0.5*33.33^2 -9.8*L*sim(15)

L =159 m