A block of mass 4 kg is kept over a rough horizontal surface. The coefficient of friction between the block and the surface is 0.1 At $t=0, \left(3\widehat{i}\right)\frac{m}{s}$ velocity is imparted to the block and simultaneously $(-2\widehat{i}) N$ force starts acting on it. Its displacement in first 5s is

$$\begin{gathered} f=\mu mg \\ \Rightarrow f=0.1\left(40\right)=4N \end{gathered}$$

frictional force will be opposing the relative motion, so the direction of the force will be along negative x-axis.

So total force acting along negative x-axis is 6N

$\mathrm{a}=\frac{-6}{4}=-1.5{\mathrm{ms}}^{-2}$

$\mathrm{t}=\frac{\mathrm{u}}{\mathrm{a}}=2\sec$

So the block will come to rest after 2sec

So the displacement of the block, $\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2=3.2+\frac{1}{2}(-1.5)2^2=3$

So the displace will be $3\hat{\mathrm{i}}$