Starting from rest, a body slides down a  $45°$  inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

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 $10 \mathrm{m}/{\mathrm{s}}^2)$

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 $g=10 \mathrm{m} {\mathrm{s}}^{-2}$

A uniform rope of total length $l$ is at rest on a table with fraction $f$ of its length hanging (see figure). If the coefficient of friction between the table and the chain is $\mu$, then

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 

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