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: $(g = 10 \ m/s^2)$

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 $g = 10 \, ms^{-2}$)

$A$ 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:

Consider a system of two masses and a pulley shown in the figure. The coefficient of friction between the two blocks and also between the bottom block and the table is $\mu = 0.1$. Find the force $F$ that must be applied to the $0.8 \text{ kg}$ block such that it attains an acceleration of $5 \text{ m/s}^2$. (Assume acceleration due to gravity, $g = 10 \text{ m/s}^2$.) (in $\text{ N}$)

The coefficient of friction between blocks $A$ and $B$ is $\mu$. The minimum force $F$ with which $A$ must be pushed such that $B$ will not slip down is

If the acceleration of block $A$ is $2 \, m/s^2$,which is provided by the frictional force from block $B$,then the magnitude of the frictional force applied by block $B$ on block $A$ is ......... $N$.