What is the maximum force $F$ that can be applied on block $m_1$,so that both $m_1$ and $m_2$ will move together? There is no friction between $m_1$ and the horizontal table. The coefficient of friction between $m_1$ and $m_2$ is $\mu$.

Two blocks $A$ and $B$ of masses $5 \,kg$ and $3 \,kg$ respectively rest on a smooth horizontal surface with $B$ over $A$. The coefficient of friction between $A$ and $B$ is $0.5$. The maximum horizontal force (in $kg \,wt.$) that can be applied to $A$,so that there will be motion of $A$ and $B$ without relative slipping,is

In the figure,the coefficient of friction between the floor and the block $B$ is $0.1$. The coefficient of friction between the blocks $B$ and $A$ is $0.2$. The mass of $A$ is $\frac{m}{2}$ and the mass of $B$ is $m$. What is the maximum horizontal force $F$ that can be applied to the block $B$ so that the two blocks move together?

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 $g = 10\,m/s^2$]

$A$ 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 $\mu_{1} / \mu_{2}$?

$A$ 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 ...... $J$.