Question

A marble block of mass $2kg$ lying on the ice, when given a velocity of $6m/s$ is stopped by friction in $10s$. Then find the value of the coefficient of friction if $g=10m/s^2$.

• A. $0.01$
• B. $0.02$
• C. $0.03$
• D. $0.06$

Answer 1

The correct option is D $0.06$

The given situation can be reframed as shown below

Let the coefficient of the friction be $\mu$
Therefore, friction force acting on block, $f=\mu mg=2\mu g$
So, from the FBD we have
$a=\frac{f}{m}=-\frac{2\mu g}{2}=-\mu g$
Since, retardation takes place here, we have taken acceleration $a$ as $-ve$
So, from the equation of motion we have
$v=u+at$
Since the blocks finally comes to rest,
$\Rightarrow 0=u-\mu gt$
$\therefore \mu =\frac{u}{gt}=\frac{6}{10\times 10}=0.06$