Starting from rest, the time taken by a body sliding down on a rough inclined plane at 45° with the horizontal is, twice the time taken to travel on a smooth plane of same inclination and same distance. Then the coefficient of kinetic friction is

Two blocks ($m=0.5 \mathrm{kg}$ and $M=4.5 \mathrm{kg}$) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is $\frac{3}{7}.$ Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is $N.$ (Round off to the Nearest Integer) [Take $g$ as $9.8 \mathrm{m} {\mathrm{s}}^{-2}$]

A brick of mass $2 \mathrm{kg}$ slides down an incline of height $5 \mathrm{m}$ and angle $30°.$ If the coefficient of friction of the incline is $\frac{1}{2\sqrt{3}},$ the velocity of the block at the bottom of the incline is (Assume the acceleration due to gravity is $10 \mathrm{m}/{\mathrm{s}}^2)$

As shown in the figure, a block of mass $\sqrt{3} \mathrm{kg}$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3\sqrt{3}}$. The critical force to be applied on the vertical surface as shown at an angle $60°$ with horizontal such that it does not move, will be $3x$. The value of $x$ will 

$\left[g=10 \mathrm{m} {\mathrm{s}}^{-2}; \sin 60°=\frac{\sqrt{3}}{2}; \cos 60°=\frac{1}{2}\right]$

The coefficient of static friction between two blocks is $0.5$ and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is _______$\mathrm{N}$ (take $\left.g=10{ \mathrm{m} \mathrm{s}}^{-2}\right)$

A block of mass $5 \mathrm{kg}$ is kept against an accelerating wedge with a wedge angle of $45°$ to the horizontal. The co-efficient of friction between the block and the wedge is $\mu =0.4$. What is the minimum absolute value of the acceleration of the wedge to keep the block steady. Assume $g=10 \mathrm{m} {\mathrm{s}}^{-2}$