Question
A block is kept on a frictionless inclined surface with angle of inclination $$'\alpha '.$$ The incline is given an acceleration $$'a'$$ to keep the block stationary. Then $$a$$ is equal to
A block is kept on a frictionless inclined surface with angle of inclination $$'\alpha '.$$ The incline is given an acceleration $$'a'$$ to keep the block stationary. Then $$a$$ is equal to
A.
$$g\,{\text{cosec}}\,\alpha $$
B.
$$\frac{g}{{\tan \alpha }}$$
C.
$$g\tan \alpha $$
D.
$$g$$
Answer :
$$g\tan \alpha $$
Solution :
From diagram,
For block to remain stationary,
$$mg\sin \alpha = ma\cos \alpha \Rightarrow a = g\tan \alpha $$
From diagram,
For block to remain stationary,
$$mg\sin \alpha = ma\cos \alpha \Rightarrow a = g\tan \alpha $$
