Question

A particle slides down a smooth inclined plane of elevation $\theta$, fixed in an elevator going up with an acceleration $a_0$. The base of the incline has a length L. Find the time taken by the particle to reach the bottom.

• A. ${\left[\frac{2L}{(g+a_0)sin\theta cos\theta }\right]}^{\frac{1}{2}}$
• B. ${\left[\frac{L}{(g+a_0)sin\theta cos\theta }\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{3L}{(g+a_0)sin\theta cos\theta }\right]}^{\frac{1}{2}}$
• D. ${\left[\frac{4L}{(g+a_0)sin\theta cos\theta }\right]}^{\frac{1}{2}}$

Answer 1

The correct option is A

${\left[\frac{2L}{(g+a_0)sin\theta cos\theta }\right]}^{\frac{1}{2}}$

Let us work in the elevator frame. A pseudo force Ma₀ in the downward direction is to be applied to the particle of mass m together with the real forces. Thus, the forces on m are
i) N normal force
ii) mg downward (by earth)
iii) $m{a}_0$ downward (pseudo)

Let a be the accelaration of the particle with respect to the incline. Taking components of the forces parallel to the incline and applying Newton’s law,
$mgsin\theta +m{a}_{0}sin\theta =ma$
or, $a=(g+a_0)sin\theta$
This is the accelaration with respect to the elevator. In this frame, the distance travelled by the particle is $\frac{L}{cos\theta }$. Hence,
$\frac{L}{cos\theta }=\frac{1}{2}(g+a_0)sin\theta .t^2$
or, $t={\left[\frac{2L}{(g+a_0)sin\theta cos\theta }\right]}^{\frac{1}{2}}$