The coefficient of friction between 4kg and 5kg blocks is 0.2 and between 5kg block, ground is 0.1 respectively. Find minimum force ‘F’ required for slipping between 4kg and 5kg blocks.

Minimum force reed to cause system to move $=\text{q}\mu \text{g}=9\text{N}$

Where force is 4 N friction between 5 kg block and grand will be 4 N system will be where at frictional force between 4 kg and 5 kg block will be zero.

If then is relative motion between 4 kg and 5 kg block

${\text{f}}_{\text{2}}=4,{\mu }_2\text{g}=4\times 0.2\times \text{g}=8\text{N}$

${\text{f}}_{\text{l}}={\mu }_1\text{g}=9\times 0.1\times \text{g}=9\text{N}$

Let acceleration of 5 kg block is ${\text{a}}_1$

$5{\text{a}}_{\text{1}}=\text{F}-\left({\text{f}}_{\text{1}}+{\text{f}}_{\text{2}}\right)$

$5{\text{a}}_{\text{1}}=\text{F}-17=1$

${\text{a}}_{\text{1}}=\frac{\text{F}-17}{5}$

${\text{F}}_{\text{pseudo}}=4\text{a}$

For relative motion between 4kg and 5 kg

$\text{5kg}{\text{f}}_{\text{pseudo}}={\text{f}}_{\text{2}}$

$4{\text{a}}_{\text{1}}\ge 8,4\times \frac{\text{F}-17}{5}\ge 8$

$\text{F}-17\ge 10$

$\text{F}\ge 27\text{N}$