Question

The retarding acceleration of $4m/s^2$ due to frictional force stops a block of mass $40kg$ moving on a rough horizontal surface. The coefficient of friction between the block and the surface is

• A. $0.3$
• B. $0.4$
• C. $0.55$
• D. $0.65$

Answer 1

The correct option is B $0.4$

As we know, coefficient of friction $\mu =\frac{f}{N}$
where $f$ is the force of friction acting on the block.

$f=ma$ for retardation $a$

$\Rightarrow \mu =\frac{ma}{mg}=\frac{a}{g}$
$a=4m{s}^{-2}\left(\text{given}\right)$
$\therefore \mu =\frac{4}{10}=0.4$