When a block moves down a smooth inclined plane of inclination $\theta$, its velocity on reaching the bottom is $v$. When it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is $v/n$, where $n$ is a number greater than $1$. The coefficient of friction between the block and the rough surface is

Consider a block kept on an inclined plane (inclined at $45°$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(\mu )$ is equal to :

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