Question

A bock of mass $m$ is placed on a frictionless inclined plane of inclination $\theta$. The inclined plane has its base fixed on the floor of a lift which is going up with a constant acceleration $a$. When the block is released, it will slide down the plane with acceleration

• A. $(g+a)\sin \theta$
• B. $(g-a)\sin \theta$
• C. $(g+a)\cos \theta$
• D. $(g-a)\cos \theta$

Answer 1

The correct option is A $(g+a)\sin \theta$

Figure shows the free body diagram of the block.
The forces acting on the block are:
(i) weight $mg$ downwards
(ii) normal reaction $N$ perpendicular to the inclined plane
(iii) reaction force ma downwards (pseudo force).
The components of $mg$ and $ma$ parallel to the inclined plane are $mg\sin \theta$ and $ma\sin \theta$ respectively. The net force on the block sliding down the plane is
$F=mg\sin \theta +ma\sin \theta =m(g+a)\sin \theta$
$\therefore$ Acceleration of the block $=\frac{F}{m}=(g+a)\sin \theta$