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

In the shown arrangement,the mass of $A = 1\,kg$ and the mass of $B = 2\,kg$. The coefficient of friction between $A$ and $B$ is $\mu = 0.2$. There is no friction between $B$ and the ground. $A$ horizontal force $F = 5\,N$ is applied to block $B$. The frictional force exerted by $A$ on $B$ is equal to:

The blocks are given velocities in the directions shown in the figure. The coefficient of friction between the two blocks is $\mu = 0.5$. Take $g = 10\ m/s^2$. (Consider that the $4\ kg$ block does not fall off the $2\ kg$ block).
Consider the following statements:
$(i)$ Time when relative motion between them stops is $1.4\ s$.
$(ii)$ Time when relative motion between them stops is $1.2\ s$.
$(iii)$ The common velocity of the two blocks is $8\ m/s$,towards the right.
$(iv)$ The displacement of the $4\ kg$ block when relative motion stops is $10.8\ m$.
Which of the statements is/are correct?

$A$ block of mass $m$ is placed on the top of another block of mass $M$ as shown in the figure. The coefficient of friction between them is $\mu$. What is the maximum acceleration with which the block $M$ may move so that $m$ also moves along with it?

When $F = 2 \text{ N}$,the frictional force between the $10 \text{ kg}$ block and the $5 \text{ kg}$ block is ...... $\text{ N}$. (Assume $g = 10 \text{ m/s}^2$)

$A$ system of two blocks of masses $m = 2 \; kg$ and $M = 8 \; kg$ is placed on a smooth table as shown in the figure. The coefficient of static friction between the two blocks is $0.5$. The maximum horizontal force $F$ that can be applied to the block of mass $M$ so that the blocks move together will be $\dots \; N$.