Question

A 6860 kg car traveling at 41 m/s is at the foot of a hill that rises 491 m in 1.7 km. At the top of the hill, the speed of the car is 5 m/s. The acceleration of gravity is 9.81 m/s 2 . Assuming constant acceleration, find the average power delivered by the car’s engine, neglecting any internal frictional losses. Answer in units of kW.

Answer 1

given

vi = 42 m/s
vf = 5 m/s
h = 491 m
d = 1.7 km = 1700 m
let a is the acceleration of the car.

a = (vf^2 - vi^2)/(2*d)

= (5^2 - 41^2)/(2*1700)

= -0.487 m/s^2

time taken to reach the top of the hill, t = (vf - vi)/a

= (5 - 41)/(-0.487)

= 73.92 s

Net workdone = change in kinetic energy

W_engine + W_gravity = (1/2)*m*(vf^2 -vi^2)

W_engine - m*g*h = (1/2)*m*(vf^2 -vi^2)

W_engine = m*g*h + (1/2)*m*(vf^2 -vi^2)

= 6860*9.81*491 + (1/2)*6860*(5^2 - 41^2)

= 2.736*10^7 J

Average power delivered by the car's engine = W_engine/t

= 2.736*10^7/73.92

= 3.70*10^5 W

= 370 kW