Question

A block B is pushed momentarily along a horizontal surface with an initial velocity V, if $\mu$ is the coefficient of sliding friction between B and the surface, the velocity of block B becomes half after time

• A. $\frac{2g\mu }{V}$
• B. $\frac{2g}{V}$
• C. $\frac{V}{2g}$
• D. $\frac{V}{\left(2g\mu \right)}$

Answer 1

The correct option is D $\frac{V}{\left(2g\mu \right)}$

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 along horizontal is friction, which will cause the retardation $a$, of the block
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:
$v=u+at$
$\Rightarrow V/2=V-at\Rightarrow t=\frac{V}{2a}=\frac{V}{2\mu g}$