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 $5 \, kg$ block and the ground is $.......... \, N$.

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

Consider the system shown below. $A$ horizontal force $F$ is applied to a block $X$ of mass $8 \,kg$,such that the block $Y$ of mass $2 \,kg$ adjacent to it does not slip downwards under gravity. There is no friction between the horizontal plane and the base of the block $X$. The coefficient of friction between the surfaces of blocks $X$ and $Y$ is $0.5$. The minimum value of $F$ is ............ $N$ (take acceleration due to gravity to be $10 \,ms^{-2}$).

$A$ $40 \,kg$ slab rests on a frictionless floor as shown in the figure. $A$ $10 \,kg$ block rests on the top of the slab. The static coefficient of friction between the block and slab is $0.60$ while the kinetic friction is $0.40$. The $10 \,kg$ block is acted upon by a horizontal force $100 \,N$. If $g = 9.8 \,m/s^2$,the resulting acceleration of the slab will be ........ $m/s^2$.

$A$ truck of mass $8 \text{ ton}$ is carrying a block of mass $2 \text{ ton}$. If a breaking force of $25 \text{ kN}$ is applied on the truck, then the frictional force acting on the block is (Coefficient of static friction between the block and the truck is $0.3$) (in $\text{ N}$)