Q. A car of mass 2000 kg is moving with a speed of 10 $ms^{- 1}\ $on a circular path of radius 20 m on a level road. What must be the frictional force between the car and the road so that the car does not slip?
A) $10^{4}\ $N
B) $10^{3}\ $N
C) $10^{5}\ $N
D) $10^{2}\ $N
Explanation
(a) To avoid slipping friction force
$F = \frac{mv^{2}}{r}$
$F = \frac{2000 \times 10 \times 10}{20} = 10^{4}\ $N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.