When $F = 2 \text{ N}$, what is the frictional force between the $5 \text{ kg}$ block and the ground (in $\text{ N}$)?

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

The coefficient of friction between two blocks is $0.2$,and the coefficient of friction between block $B$ and the surface is $0.5$. If a force of $25 \ N$ is applied to block $B$ (mass $A = 2 \ kg$,mass $B = 8 \ kg$),what is the magnitude of the frictional force between the two blocks in $N$?

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:

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

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 $\mu$.