Question

A car of mass $2000kg$ is moving with a constant speed of $10m{s}^{-1}$ on a circular path of radius $20\text{m}$ on a level road. What must be the frictional force between the car and the road so that the car does not slip on the road ?

• A. ${10}^4\text{N}$
• B. ${10}^3\text{N}$
• C. ${10}^5\text{N}$
• D. ${10}^2\text{N}$

Answer 1

The correct option is A ${10}^4\text{N}$

To avoid slipping, frictional force will act in a direction to provide the necessary centripetal force. i.e frictional force will act towards centre of circular curvature.
Here,$f$ is magnitude of frictional force.
$\therefore$ $f=\frac{m{v}^2}{r}$
$f=\frac{2000\times {10}^2}{20}={10}^4\text{N}$