The frictional force acting on the $1 \,kg$ block is .................. $N$.

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

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:

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

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 $\mu$.

When $F = 2 \, N$,the frictional force between the $5 \, kg$ block and the ground is $.......... \, N$.