Q. A brick of mass $2\ kg$ begins to slide down on a plane inclined at an angle of $45{^\circ}$ with the horizontal. The force of friction will be
A) $19.6\sin{45{^\circ}}$
B) $19.6\cos{45{^\circ}}$
C) $9.8\sin{45{^\circ}}$
D) $9.8\cos{45{^\circ}}$
Explanation
(a) For angle of repose,
Friction = Component of weight along the plane
$= mg\sin\theta = 2 \times 9.8 \times \sin{45{^\circ}} = 19.6\sin{45{^\circ}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.