Problema Solution
A 2 ton truck descending a 5° hill is brought to a stop in 250ft. The driver applies the brakes so that the wheels lock. a) If the coefficient of kinetic friction between the truck tires and the road is 0.60, how much work is done by friction in stopping the truck? b) How much work is done on the truck by gravity? how fast was the truck travelling immediately before the brakes were applied?
Answer provided by our tutors
2 ton = 2000 kg
g = 9.8 m/s^2
250 ft = 76.2 m
(a)
Work done by friction in stopping the truck = f*s = 0.60*2000kg*9.8m/s^2*cos(5)*76.2 = 8.93*10^5 J
(b)
Work done on the truck by gravity = mgh = 2000kg*9.8m/s^2*76.2*sin(5) = 1.30*10^5 J
(c)
The speed before the brakes were applied is v
1/2 mv^2 = E2-E1
1/2 *2000 kg*v^2 = 8.93*10^5 J - 1.30*10^5 J
v = 27.6 m/s
So the speed before the brakes were applied is 27.6 m/s