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

Two blocks $(A)$ of $2\,kg$ and $(B)$ of $5\,kg$ rest one over the other on a smooth horizontal plane. The coefficient of static and dynamic friction between $(A)$ and $(B)$ is the same and equal to $0.60$. Find the maximum horizontal force that can be applied to $(B)$ such that both $(A)$ and $(B)$ do not have any relative motion. $(g = 10\,m/s^2)$

The coefficient of static friction between two blocks is $0.5$ and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is $\ldots \ldots N$. (take $g=10 \, m/s^2$)

$A$ $40 \,kg$ 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 ............. $m/s^2$.

Two blocks $A$ and $B$ of masses $4 \ kg$ and $6 \ kg$ are as shown in the figure. $A$ horizontal force of $12 \ N$ is required to make $A$ slip over $B$. Find the maximum horizontal force $F_B$ that can be applied on $B$ so that both $A$ and $B$ move together. (Take $g=10 \ m \ s^{-2}$) (in $N$)

$A$ block of mass $M$ is placed on a horizontal surface and is tied with an inextensible string to a block of mass $m$,as shown in the figure. $A$ block of mass $m_0$ is also placed on $M$. If friction exists between the block $M$ and the block $m_0$ and there is no friction between the block $M$ and the horizontal surface,then the minimum value of $\mu$ for which the block $m_0$ remains stationary with respect to $M$ is: