Question

The retarding acceleration of $7.35{\text{ms}}^{-2}$ due to frictional force stops the car of mass $400\text{kg}$ travelling on a 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 $0.75$

Using Newtons 2nd law
$F_{\text{net}}=ma$
$\mu mg=ma$
$\Rightarrow \mu =\frac{ma}{mg}=\frac{a}{g}$
($a=7.35{\text{ms}}^{-2}$ given)
$\therefore \mu =\frac{7.35}{9.8}=0.75$