$A$ force $\vec{F}=\hat{i}+4 \hat{j}$ acts on the block shown in the figure. The force of friction acting on the block is (take $g=10 \, m/s^2$)

$A$ conveyor belt is moving horizontally with a velocity of $2 \,m \,s^{-1}$. If a body of mass $10 \,kg$ is kept on it, then the distance travelled by the body before coming to rest relative to the belt is (The coefficient of kinetic friction between the belt and the body is $0.2$ and acceleration due to gravity is $10 \,m \,s^{-2}$) (in $\,m$)

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move (in $N$)? (Given $m = \sqrt{3} \ kg$,$\mu = \frac{1}{2\sqrt{3}}$,$g = 10 \ m/s^2$)

$A$ block of mass $10 \ kg$ is moving on a rough surface with a velocity $v = 4 \ m/s$ to the left. An external force $F = 20 \ N$ is applied to the right as shown in the figure. The coefficient of kinetic friction is $\mu = 0.6$. The frictional force acting on the block is ...... $N$.

$A$ block of mass $5 \ kg$ is placed on a rough horizontal surface having a coefficient of friction $0.5$. If a horizontal force of $60 \ N$ is acting on it,then the acceleration of the block is (Acceleration due to gravity,$g = 10 \ ms^{-2}$). (in $ms^{-2}$)

$A$ particle of mass '$m$' at rest on a rough horizontal surface with a coefficient of friction '$\mu$' is given a velocity '$u$'. The average power imparted by friction before it stops is: