Question

A car of mass $1500$ kg is moving with a speed of $12.5$ m/s on a circular path of radius $20$ m on a level road. What should be the coefficient of friction between the car and the road, so that the car does not slip?

• A. $0.2$
• B. $0.4$
• C. $0.6$
• D. $0.8$

Answer 1

Answer: (D)

$0.8$

Given : Radius of the circular path $r=20$ $m$

Mass of the car $M=1500kg$

Speed $v=12.5$ $m/s$

Let the coefficient of friction be $\mu$.

The centrifugal force will balance the friction force so that the car does not slip.

$\therefore$ $\frac{m{v}^2}{r}=\mu mg$ $⟹$ $\mu =\frac{v^2}{rg}$

$\mu =\frac{(12.5{)}^2}{20\times 9.8}=0.8$