A box of mass $6 kg$ rests upon an inclined plane. The inclination of the plane to the horizontal direction is gradually increased. It is found that when the slope of the plane is $2$ in $3$, the box starts sliding down the plane. Find the co-efficient of friction between the box and the plane. What force applied to the box parallel to the plane will make it move up the plane ? $g = 9.8 m{s}^{-2}$.

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 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 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 $10 \mathrm{m}/{\mathrm{s}}^2)$