$A$ body of mass $1 \,kg$ is moving with a velocity $10 \,ms^{-1}$ on a horizontal rough surface having a coefficient of kinetic friction $0.4$. If the constant force is removed, the body comes to rest in a time (Acceleration due to gravity $g = 10 \,ms^{-2}$). (in $\,s$)

$A$ particle of mass $M$ moves along a horizontal $x$-axis from $x=0$ to $x=L$. The coefficient of kinetic friction varies as a function of $x$ as $\mu(x) = \mu_0 - \alpha x$, where $\mu_0$ and $\alpha$ are constants of appropriate dimensions, such that $\mu(L)=0$. The total work done by the frictional force during the motion is $n\mu_0 MgL$, where $g$ is the acceleration due to gravity. The value of $n$ is:

What is the angle between the instantaneous velocity and the friction force of an object moving on a rough surface (in $^{\circ}$)?

$A$ 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 $w$?

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 ........ $m$. $[g = 10 \, m/s^2]$

$A$ stationary body of mass $m$ is slowly lowered onto a massive platform of mass $M$ $(M \gg m)$ moving at a speed $V_p = 4 \, m/s$ as shown in the figure. How far will the body slide along the platform (in $, m$)? (Given: $\mu = 0.2$ and $g = 10 \, m/s^2$)