$A$ pen of mass $m$ is lying on a piece of paper of mass $M$ placed on a rough table. If the coefficients of friction between the pen and paper and the paper and the table are $\mu_1$ and $\mu_2$,respectively,then the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by:

Two blocks $A$ and $B$ of masses $m_A = 1\,kg$ and $m_B = 3\,kg$ are kept on a table as shown in the figure. The coefficient of friction between $A$ and $B$ is $\mu_1 = 0.2$ and between $B$ and the surface of the table is $\mu_2 = 0.2$. The maximum horizontal force $F$ that can be applied on $B$ such that block $A$ does not slide over block $B$ is ........ $N$. [Take $g = 10\,m/s^2$]

$A$ $30 \ kg$ slab $B$ rests on a frictionless floor as shown in the figure. $A$ $10 \ kg$ block $A$ rests on top of the slab $B$. The coefficients of static and kinetic friction between the block $A$ and the slab $B$ are $0.60$ and $0.40$ respectively. When block $A$ is acted upon by a horizontal force of $100 \ N$ as shown,find the resulting acceleration of the slab $B$. $(g = 9.8 \ m \ s^{-2})$ (in $m \ s^{-2}$)

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

The value of $\theta$ is increased gradually from $\theta = 0$. At $\theta = \tan^{-1}(1/2)$,both blocks just start slipping. Then the value of $\mu_2$ is: $(g = 10 \ m/s^2)$

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).