Question

A car is moving along a straight horizontal road with a speed $v_{0}$. If the coefficient of friction between the tyres and the road is $\mu$, the shortest distance in which the car be stopped is :

• A. $\frac{v_{0}^{2}}{\mu}$
• B. $\left(\frac{v_{0}}{\mu_{g}}\right)^{2}$
• C. $\frac{v_{0}^{2}}{\mu_{g}}$
• D. $\frac{v_{0}^{2}}{2 \mu_{g}}$

Answer 1

Answer: (D)

Work dorce against frictional force equals the kinetic energy uf the body.
When a body of mass $m$, mioves with velocity $v$,
it has kinetic energy $K=\frac{1}{2} m v^{2}$,
this energy is utilized in doing work against the frictional force between the tyres of the car and road.
$\therefore $ Kinetic energy $=$ Work done against friction force
$\frac{1}{2} m v^{2}=\mu m g s$
where $s$ is the distance in which the car is stopped and $\mu$ is coefficient of kinetic friction.
Given $v=v_{0} $
$\therefore s=\frac{v_{0}^{2}}{2 \mu g}$