$A$ 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 $\text{ N}$)

$A$ 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:

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

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

$A$ $40 \,kg$ 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 ........ $m/s^2$.

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 $\text{ N}$)