$A$ vehicle of mass $M$ is moving with momentum $P$ on a rough horizontal road. The coefficient of friction between the tyres and the horizontal road is $\mu$. The stopping distance is ($g$ $=$ acceleration due to gravity).

$A$ conveyor belt is moving at a constant speed of $2\, m s^{-1}$. $A$ box is gently dropped on it. The coefficient of friction between them is $\mu = 0.5$. The distance that the box will move relative to the belt before coming to rest on it,taking $g = 10\, m s^{-2}$,is ........... $m$.

$A$ block of mass $m$ (initially at rest) is sliding up a rough vertical wall with the help of a force $F$ whose magnitude is constant but direction is changing. $\theta = {\theta _0}t$,where $t$ is time in seconds. At $t = 0$,the force is in the vertical upward direction,and as time passes,its direction changes until $\theta = \frac{\pi }{2}$. The value of $F$ such that the block comes to rest when $\theta = \frac{\pi }{2}$ is:

$A$ uniform rope of total length $l$ is at rest on a table with a fraction $f$ of its length hanging (see figure). If the coefficient of friction between the table and the rope is $\mu$,then:

$A$ 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 $\,m$)

$A$ car is moving on a straight horizontal road with a speed $v.$ If the coefficient of friction between the tyres and the road is $\mu ,$ the shortest distance in which the car can be stopped is