Two blocks,$4\, kg$ and $2\, kg$,are sliding down an inclined plane as shown in the figure. The acceleration of the $2\, kg$ block is ........ $m/s^2$.

In the given figure, the blocks $A$, $B$, and $C$ weigh $4 \ kg$, $6 \ kg$, and $8 \ kg$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is . . . . . . $N$. (Use $g = 10 \ m/s^2$)

$A$ man can move on a horizontal plank supported symmetrically as shown. The variation of normal reaction on support $A$ with distance $x$ of the man from the end of the plank is best represented by:

$A$ small block of mass $0.1 \ kg$ lies on a fixed inclined plane $PQ$ which makes an angle $\theta$ with the horizontal. $A$ horizontal force of $1 \ N$ acts on the block through its center of mass as shown in the figure. The block remains stationary if (take $g = 10 \ m/s^2$):

$A$ truck starting from rest moves with an acceleration of $5\,m/s^2$ for $1\,sec$ and then moves with constant velocity. The velocity w.r.t. ground v/s time graph for the block in the truck is (Assume that the block does not fall off the truck and the coefficient of friction $\mu = 0.2$):

$A$ hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is