For the given situation shown in figure, choose the correct options (Take, $g=10{ \mathrm{ms}}^{-2}$ )

 $f_1\to$ force of friction between 2kg and $4 \mathrm{kg}$

 $f_2\to$ force of friction between 4kg and ground

$$\begin{gathered} {\left(f_{S_1}\right)}_{\max }=0.4\times 2\times 10=8 \mathrm{N} \\ {F}_{K_1}=0.2\times 2\times 10=4 \mathrm{N} \\ {\left(f_{S_2}\right)}_{\max }=0.6\times 6\times 10=36 \mathrm{N} \\ {F}_{K_2}=0.4\times 6\times 10=24 \mathrm{N} \end{gathered}$$

At $t=1 \mathrm{s},F=2 \mathrm{N}<36 \mathrm{N}$, therefore system remains stationary and force of friction between $(2 \mathrm{kg})$ and 4kg is zero.

At $t=4 \mathrm{s},F=8 \mathrm{N}<36 \mathrm{N}$. Therefore system is again stationary and force of friction on 4kg from ground is 8N.

At $t=15 \mathrm{s},F=30 \mathrm{N}<36 \mathrm{N}$and system is stationary.