In the figure,the coefficient of friction between the floor and block $B$ is $\mu_1 = 0.2$ and between blocks $A$ and $B$ is $\mu_2 = 0.3$. The masses are $m_A = 20 \text{ kg}$ and $m_B = 40 \text{ kg}$. Find the maximum horizontal force $F$ that can be applied to block $B$ so that both blocks move together. (in $\text{ N}$)

When $F = 2 \text{ N}$, what is the frictional force between the $5 \text{ kg}$ block and the ground (in $\text{ N}$)?

Block $A$ of mass $2 \, kg$ is placed over block $B$ of mass $8 \, kg$. The combination is placed over a rough horizontal surface. The coefficient of friction between block $B$ and the floor is $0.5$. The coefficient of friction between blocks $A$ and $B$ is $0.4$. A horizontal force of $10 \, N$ is applied on block $B$. Find the force of friction between blocks $A$ and $B$. (Take $g = 10 \, ms^{-2}$)

In the given arrangement,find the maximum value of $F$ for which there is no relative motion between the blocks.

$A$ $40 \,kg$ slab rests on a frictionless floor as shown in the figure. $A$ $10 \,kg$ block rests on the top of the slab. The static coefficient of friction between the block and slab is $0.60$ while the kinetic friction is $0.40$. The $10 \,kg$ block is acted upon by a horizontal force $100 \,N$. If $g = 9.8 \,m/s^2$,the resulting acceleration of the slab will be ........ $m/s^2$.

When $F = 2\,N$,the frictional force between the $10\,kg$ block and the $5\,kg$ block is $..........\,N$. (Given: $\mu_s = 0.1$ between blocks,$\mu_s = 0.3$ between $5\,kg$ block and ground).