Question

The driver of a 1550 kg car, initially traveling at 12.1 m/s, applies the brakes, bringing the car to rest in a distance of 18.5 m.

(a) Find the net work done on the car.

(b) Find the magnitude and direction of the force that does this work. (Assume this force is constant.)

Answer 1

a)
work done = change in kinetic energy
= 0.5*m*(Vf^2 - Vi^2)
where, Vf is fianl velocity that is 0
so,
work done = 0.5*1550*(0 - (12.1)^2)
= -0.5*1550*146.41
= -113467.75 J

since, work is scalar quantity
so,
work done = 113467.75 J
= 113.5 KJ

b)
use,
work done = F*d
113467.75 = F*18.5
F = 6133.39 N
= 6.13 kN

since, velocity decreases
so, force will be in opposite direction of motion of car
so,
F = -6.13 kN