$A$ force of $0.5 \ N$ is applied on the upper block as shown in the figure. The coefficient of static friction between the two blocks is $0.1$ and that between the lower block and the surface is zero. The work done by the lower block on the upper block for a displacement of $3 \ m$ of the upper block is ...... $J$.

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 $\text{ N}$)

$A$ block of mass $m_{2}$ is placed on a horizontal table and another block of mass $m_{1}$ is placed on top of it. An increasing horizontal force $F=\alpha t$ is exerted on the upper block, but the lower block never moves as a result. If the coefficient of friction between the blocks is $\mu_{1}$ and that between the lower block and the table is $\mu_{2}$, then what is the maximum possible value of $\mu_{1} / \mu_{2}$?

Find the maximum force $F$ (in $N$) which will not cause motion of any of the blocks as shown in the figure.

The system shown in the figure is released. Find the accelerations $a_1$ (for the $2 \ kg$ block) and $a_2$ (for the $3 \ kg$ block).

If $100 \,N$ force is applied to $10 \,kg$ block as shown in the diagram,the acceleration of $40 \,kg$ slab is (in $\,ms^{-2}$)