For the given figure,find the acceleration of the $40\, kg$ block in $m/s^2$. (Take $g = 10\, m/s^2$)

$A$ 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:

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

$A$ cylinder of mass $12 \,kg$ is sliding on a plane with an initial velocity $20 \,ms^{-1}$. If the coefficient of friction between the surface and the cylinder is $0.5$, what is the distance the cylinder travels before stopping (in $\,m$)?

$A$ block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in the figure. The coefficient of kinetic friction is $\mu_{K}$. Then,the block's acceleration $a$ is given by: ($g$ is acceleration due to gravity)

$A$ fireman of mass $60 \ kg$ slides down a pole. He is pressing the pole with a force of $600 \ N$. The coefficient of friction between the hands and the pole is $0.5$. With what acceleration (in $m/s^2$) will the fireman slide down? (Take $g = 10 \ m/s^2$)