When a body slides down an inclined plane with a coefficient of friction $\mu$,its acceleration will be:

$STATEMENT-1$: $A$ block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30^{\circ}$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. because
$STATEMENT-2$: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

Two blocks $A$ and $B$ of equal masses are sliding down along straight parallel lines on an inclined plane of $45^{\circ}$. Their coefficients of kinetic friction are $\mu_A = 0.2$ and $\mu_B = 0.3$ respectively. At $t = 0$,both the blocks are at rest and block $A$ is $\sqrt{2} \text{ m}$ behind block $B$. Calculate the time and distance from the initial position where the front faces of the blocks come in line on the inclined plane as shown in the figure. (Use $g = 10 \text{ m/s}^2$)

$A$ body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $\mu$. The work done against friction before the body comes to rest is

Two blocks of masses $1 \ kg$ and $2 \ kg$ are connected by a light rod and the system is slipping down a rough incline at an angle of $45^{\circ}$ with the horizontal. The coefficient of kinetic friction at both contacts is $0.4$. If the acceleration of the system is $\alpha \sqrt{2} \ m/s^2$,find the value of $\alpha$. (Use $g = 10 \ m/s^2$)

$A$ body is sliding down an inclined plane (angle of inclination $45^{\circ}$). If the coefficient of friction is $0.5$ and $g = 9.8\, m/s^2$,then the acceleration of the body downwards in $m/s^2$ is