$A$ man slides down a telegraphic pole with an acceleration equal to one-fourth of the acceleration due to gravity. The frictional force between the man and the pole is equal to which of the following in terms of the man's weight $w$?

$A$ block of mass $m$ is moving on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is $\mu_{k}$. The net force exerted by the surface on the block is (where $g$ is the acceleration due to gravity).

$A$ 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 $\,m$)

$A$ 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:

$A$ block of mass $100 \ kg$ slides over a distance of $10 \ m$ on a horizontal surface. If the coefficient of friction between the surfaces is $0.4$,then the work done against friction (in $J$) is:

$A$ movable steel plate is placed between fixed steel and brass plates and the stack of plates is subjected to a weight of $100 \ N$ as shown in the figure. The coefficient of kinetic friction for steel on steel is $0.57$ and for steel on brass is $0.44$. Assuming that the entire weight comes onto the stack and that the weight of the plates is negligible in comparison to the applied weight,the force required to move the middle plate (in $N$) is