A block is kept on a frictionless inclined surface with angle of inclination $\alpha$. The incline is given…

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

1. $\mathrm{g} / \tan \alpha$
2. $\mathrm{g} \operatorname{cosec} \alpha$
3. g
4. $\mathrm{g} \operatorname{tan} \alpha$

Solution

$m g \sin \alpha=m a \cos \alpha$ $\therefore \mathrm{a}=\mathrm{g} \tan \alpha$

Asked in: JEE Main 2005

Practice more Laws of Motion questions on Aicharya