Question

The retarding acceleration of $7.35m/s^2$ due to frictioanl forcer stops the car of mass $400kg$ travelling on road. The coefficient of friction between the tyre of the car and the road is

• A. $0.55$
• B. $0.75$
• C. $0.70$
• D. $0.65$

Answer 1

The correct option is B $0.75$

As we know, coefficient of friction $\mu =\frac{F}{N}$

$\Rightarrow \mu =\frac{ma}{mg}=\frac{a}{g}$
$\mu =\frac{7.35}{9.8}=0.75$