Question

A 2000 kg truck travelling at 54km/hr on the application of break it comes to rest by covering 37.5 metres.calculate average retarding force of break

Answer 1

The magnitude of acceleration of the truck

$$a=\frac{v^2}{2d}=\frac{(54\times\frac{1000\:\rm{m}}{3600\:\rm{s}})^2}{2\times 37.5\:\rm{m}}=3\:\rm{m/s^2}$$

The average force

$$F=ma=2000\:\rm{kg}\times 3\:\rm{m/s^2}=6000\;\rm{N}$$