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?
3.In problem 1, how fast was the car travelling initially?
Answer provided by our tutors
(1)
The friction force = 0.60*2000*32.15*cos(5) = 38433N
The work done by friction = 38433N*250 = 9.61*10^6 J
(2)
The work done by gravity = mgh*sin(5) = 2000*32.15*sin(5)*250 = 1.40 *10^6 J
(3)
The initially speed is v
1/2 mv^2 = 9.61*10^6 J - 1.40 *10^6 J
v = 90.6 m/s
So the initially speed is 90.6 m/s