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An inclined plane ABCD is shown in the figure. It is inclined at an angle of $45^{0}$ from horizontal. A block is placed on the inclined plane. The plane is accelerating along x-axis with an acceleration, $a = 5 m / s^{2} .$ The minimum value of coefficient of friction between the block and the plane required, so that the block does not slip on the plane, is $(g = 10 m / s^{2})$

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$\sqrt{1.5}$

answer is A.

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Detailed Solution

$μ N \geq \sqrt{{(m g \sin θ)}^{2} + {(m a)}^{2}} \Rightarrow μ m g \cos θ \geq m \sqrt{a^{2} + g^{2} \sin^{2} θ} \Rightarrow μ \geq \sqrt{\tan^{2} θ + {(\frac{a}{g})}^{2} \sec^{2} θ}$

By putting a = g/2, and $θ = 45^{0}$

$μ = \sqrt{\frac{3}{2}}$

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