Question

A block $B$ is pushed momentarily along the horizontal surface with an initial velocity $v$. If $\mu$ is the coefficient of sliding friction between $B$ and the surface, block $B$ will come to rest after a time

• A. $\frac{v}{\mu g}$
• B. $\frac{v}{g}$
• C. $\frac{\mu g}{v}$
• D. $\frac{g}{\mu v}$

Answer 1

The correct option is A $\frac{v}{\mu g}$

Let the mass of the block be $m$
Given, initial velocity of the block is $v$ and coefficient of friction between block and the surface is $\mu$
The given situation can be shown as below

Since, the only acting force on the block is friction which will cause the retardation $a$, of the block and it will come to rest.
So, from the FBD we have
$N=mg$
$f=\mu N=\mu mg$
The retardation of the block is given by
$a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g$

Now, from equation of motion, time taken by the block to come to the rest is given by
$v=u+at$
$\Rightarrow 0=v-at\Rightarrow t=\frac{v}{a}=\frac{v}{\mu g}$