A body of mass m is placed over a smooth inclined plane of inclination $θ$ , which is placed over a lift which is moving up with an acceleration $a_{0}$ . Base length of the inclined plane is L. Calculate the velocity of the block with respect to lift at the bottom, if it is allowed to slide down from the top of the plane from rest.

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$\sqrt{2 (a_{0} + g) L \cos θ}$

$\sqrt{2 (a_{0} + g) L c o t θ}$

$\sqrt{2 (a_{0} + g) L \sin θ}$

$\sqrt{2 (a_{0} + g) L \tan θ}$

answer is C.

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Detailed Solution

Acceleration along the plane with respect to lift is,

$a = (a_{0} + g) \sin θ$

Initial velocity = 0

$v^{2} = u^{2} + 2 a s$

v = $\sqrt{2 (a_{0} + g) \sin θ \frac{L}{\cos θ}}$

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