OnlineCBT
Physics · LAWS OF MOTION
हिन्दीA block of mass \( 2\text{ kg} \) rests on an inclined plane making an angle of \( 30^\circ \) with the horizontal. The coefficient of static friction between the block and the incline is \( \mu_s = 0.6 \). Taking \( g = 10\text{ m/s}^2 \), the magnitude of the frictional force acting on the block is:
Correct answer · Option B
\( 10\text{ N} \)
Explanation
Component of weight down the incline: \( mg \sin 30^\circ = 2 \times 10 \times 0.5 = 10\text{ N} \). Maximum static friction available: \( f_{s,\text{max}} = \mu_s mg \cos 30^\circ = 0.6 \times 20 \times \frac{\sqrt{3}}{2} \approx 10.39\text{ N} \). Since \( mg \sin 30^\circ < f_{s,\text{max}} \), block remains at rest and static friction force equals \( 10\text{ N} \).
