Q. A block of mass $10\ kg$ is placed on an inclined plane. When the angle of inclination is $30{^\circ}$, the block just begin to slide down the plane. The force of static friction is
A) $10\ kg\ wt$
B) $89\ kg\ wt$
C) $49\ kg\ wt$
D) $5\ kg\ wt$
Explanation
(d) $F = mg\sin{30{^\circ}} = 50\ N = 5kg - wt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.