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Q.

A block is kept inside an elevator which is ascending up at an angle 53⁰ with the horizontal with an acceleration of $5 m / s^{2} .$ Consider $g = 10 m / s^{2},$ the minimum value of coefficient of friction for which the block does not slip on the elevator floor is $μ .$ Find the value of $14 \times μ .$

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Detailed Solution

Assuming the frictional force to act at its maximum,
$N - m (10 m / s^{2}) = m (5 m / s^{2}) \sin 53^{o}$
And $μ N = m (5 m / s^{2}) \cos 53^{o}$ we get $μ = \frac{3}{14}$

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A block is kept inside an elevator which is ascending up at an angle 530 with the horizontal with an acceleration of 5 m/s2.Consider g=10m/s2, the minimum value of coefficient of friction for which the block does not slip on the elevator floor is μ. Find the value of 14×μ.