Question

A marble block of mass $2\text{kg}$ lying on ice when given a velocity of $6\text{m/s}$ is stopped by friction in $10\text{s}$. Then the coefficient of friction is

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

Answer 1

The correct option is D $0.06$

by Newton's law equation the external force $F_{ext}=ma$.
Here the extrenal force is frictional force only and let's assume it is in opposite direction of friction force.
Hence,
$-\mu N=ma$

$a=\frac{-\mu N}{m}=\frac{-\mu mg}{m}=-\mu g$

Now using $v=u+at$, $t=\text{10 s}$
we get,
$0=6-(\mu \times 10)\times 10$, since $(a=-\mu g)$
Which gives $\mu =0.06$