A wedge $Y$ with mass of 10 kg and all frictionless surfaces and the inclined surface making $37^{\circ}$ with horizontal. A block $X$ with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At $t=0$ wedge ( $Y$ ) is pulled toward right with constant force $(f)$ of 24 N . Taking the block $X$ at rest at $t=0$, the time taken by it to slide down 8.8 m on the slope, while $Y$ is on the move, is $\_\_\_\_$ s.

$\left(\right.$ take $\tan \left(37^{\circ}\right)=3 / 4$ and $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

The time taken by a block of mass $m$ to slide down from the highest point to the lowest point on a rough inclined plane is $50 \%$ more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at $45^{\circ}$ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is $\_\_\_\_$

A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration $a_0$. The angle between the inclined plane and ground is $\theta$ and its base length is $L$. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ________.

A block of mass 5 kg is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. Friction coefficient between the block and inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is $\_\_\_\_$ N.

$$ \left(g=10 \mathrm{~m} / \mathrm{s}^2\right) . $$