For the situation shown in figure, mark the correct options.

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

$=0.4\times 2\times 10=8 \mathrm{N}$

2kg moves due to friction. Therefore its maximum acceleration may be

$a_{\max }=\frac{8}{2}=4 \mathrm{m}/{\mathrm{s}}^2$

Slip will start when their combined acceleration becomes $4 \mathrm{m}/{\mathrm{s}}^2$

$\therefore a=\frac{F}{m}$ or $4=\frac{2t}{6}$ or $t=12 \mathrm{s}$

$\mathrm{At} t=3 \mathrm{s}$

$a_2=a_4=\frac{F}{m}=\frac{2t}{6}=\frac{2\times 3}{6}=1 \mathrm{m}/{\mathrm{s}}^2$

Both $a_2$ and $a_4$ are towards right. Therefore, pseudo forces $F_1$ on $2 \mathrm{kg}$ from 4kg and $F_2$ on 4kg from 2kg are towards left

$F_1=\left(2\right)\left(1\right)=2 \mathrm{N}\Rightarrow F_2=\left(4\right)\left(1\right)=4 \mathrm{N}$

From here, we can see that $F_1$ and $F_2$ do not make a pair of equal and opposite forces.