Question

A car travels on level ground with a speed of 20m/s when it approaches a hill.

a) What vertical height does the car reach assuming the engine of the car is not engaged, and no energy is lost due to drag ?

b) After reaching the highest point, the car drives backwards towards its starting point with its brakes partially engaged, resulting to a 20 % energy loss before it reaches its initial starting point. What is the speed of the car when it returns to the initial starting point?

Answer 1

(a) Suppose mass of the car = m

Given that the initial speed of the car, v = 20 m/s

Total initial energy of the car, E = Kinetic energy of the car = (1/2)*m*v^2

Suppose the car reaches to a height of 'h' meter above the hill.

So, final energy of the car = Potential energy of the car = m*g*h

Apply conservation of energy -

Total initial energy of the car = Total final energy of the car

=> (1/2)*m*v^2 = m*g*h

=> h = v^2 / (2*g)

Put the values -

h = 20^2 / (2*9.81) = 20.4 m (Answer)

(b) In the returning journey, 20% energy lost due to brakes partially engaged.

so,

total final energy = 0.8*m*g*h

Suppose the speed of the car at the starting point in this case = v'

So, kinetic energy of the car at the starting point = (1/2)*m*v'^2

Again apply conservation of energy -

0.8*m*g*h = (1/2)*m*v'^2

=> v'^2 = 2*0.8*g*h = 2*0.8*9.81*20.4

v' = sqrt(2*0.8*9.81*20.4) = 17.9 m/s (Answer)