$A$ horizontal force of $4\,N$ is needed to keep a block of mass $0.5\,kg$ sliding on a horizontal surface with a constant speed. The coefficient of sliding friction must be: $[g = 10\,m/s^2]$

$A$ 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 $\,m$)

$A$ 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:

Consider a car moving along a straight horizontal road with a speed of $72 \, km/h$. If the coefficient of kinetic friction between the tyres and the road is $0.5$,the shortest distance in which the car can be stopped is ........ $m$. $[g = 10 \, m/s^2]$

$A$ 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).

$A$ block of mass $m$ is pulled along a horizontal surface by applying a force at an angle $\theta$ with the horizontal. The friction coefficient between the block and the surface is $\mu$. If the block travels at a uniform velocity,then the work done during its displacement $d$ is: