A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_{0}.\) The angle between the inclined plane and ground is \(\theta\) and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ___________.

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Answer & Solution

Correct Answer

(B) \(\sqrt{\frac{4L}{g~sin~2\theta-a_{0}(1+cos~2\theta)}}\)

Step-by-step Solution

Detailed explanation

\(mgsin\theta + ma_{0}cos\theta=ma\)
\(a= g~sin\theta + a_{0}cos\theta\)
Now using,
\(S=ut+\frac{1}{2}a_{down}t^{2}\)
\(\frac{L}{cos~\theta}=\frac{1}{2}(g~sin~\theta-a_{0}cos~\theta)t^{2}\)
\(t=\sqrt{\frac{2L}{g~sin\theta cos\theta-a_{0}cos^{2}\theta}}\)
\(t=\sqrt{\frac{4L}{g~sin~2\theta-a_{0}(1+cos~2\theta)}}\)

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