A bullet of mass 0 . 1   kg moving horizontally with speed 400   m   s - 1 hits a wooden…

Let the mass of the bullet be $m$ and mass of the block be $M$. It is given that $V=400 \mathrm{m} {\mathrm{s}}^{-1}$. By conservation of momentum,

$mV=(m+M)v$

 $\Rightarrow 0.1\times 400=(0.1+3.9)v$
$\Rightarrow v=10 \mathrm{m} {\mathrm{s}}^{-1}$

The force acting on the block 

$$\begin{gathered} f=\mu m\text{'}g \\ \Rightarrow a=\mu g \end{gathered}$$

Using  $v{\text{'}}^2-v^2=-2aS$

$\Rightarrow$ $0^2-{10}^2=2(-\mu g)\left(S\right)$
$\Rightarrow \mu =\frac{100}{2\times 10\times 20}=0.25$