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?
Answer provided by our tutors
250 ft = 76.2 m
(1)
The friction force = 0.60*2000*9.8*cos(5) = 11715N
The work done by friction = 11715N*76.2 m = 8.93*10^5 J
(2)
The work done by gravity = mgh*sin(5) = 2000*9.8*sin(5)*76.2 = 1.30 *10^5 J