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A small block is sliding on a frictionless inclined plane that is moving upward with a constant acceleration.If the block remains at a level height,what is the acceleration of the inclined plane? Acceleration due to gravity is $g$

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$g \tan θ$

$g c o t θ$

$g \tan^{2} θ$

$g \sin^{2} θ$

answer is D.

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Detailed Solution

Acceleration of block in vertical direction

$a_{y} = a - a^{'} \sin θ = 0$

$a - (g + a) \sin^{2} θ = 0 (g + a) \sin^{2} θ = a (1 - \sin^{2} θ) = g \sin^{2} θ a = \frac{g \sin^{2} θ}{1 - \sin^{2} θ} = \frac{g \sin^{2} θ}{\cos^{2} θ} = g \tan^{2} θ$

Hence the correct option is 4

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