Question

The coefficient of friction of an inclined plane $1/\sqrt{3}$. If it is inclined at an angle ${30}^{\circ }$ with the horizontal, what will be the downward acceleration of the block placed on the inclined plane?

• A. $0$
• B. $\sqrt{2}{ms}^{-2}$
• C. $\sqrt{3}{ms}^{-2}$
• D. $3{ms}^{-2}$

Answer 1

Answer: (A)

$0$

Let $a$ be the downward acceleration of block. From FBD,
$ma=mg\sin \theta -f=mg\sin \theta -\mu mg\cos \theta$
or $a=gsin30-\frac{1}{\sqrt{3}}g\cos 30=\frac{g}{2}-g(\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{2})$
$\therefore a=0$