Q. The coefficient of friction between a body and the surface of an inclined plane at $45{^\circ}$ is 0.5. If $g = 9.8\ m/s^{2}$, the acceleration of the body downwards in $m/s^{2}$ is
A) $\frac{4.9}{\sqrt{2}}$
B) $4.9\ \sqrt{2}$
C) $19.6\ \sqrt{2}$
D) 4.9
Explanation
(a) $a = g\left( \sin\theta - \mu\cos\theta \right) = 9.8\ (\sin{45{^\circ}} - 0.5\cos{45{^\circ}})$
$= \frac{4.9}{\sqrt{2}}\ m/s{ec}^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.