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Physics · LAWS OF MOTION
हिन्दीA block of mass \( m = 5\text{ kg} \) is placed on an inclined plane of inclination \( \theta = 37^\circ \) (\( \sin 37^\circ = 0.6, \cos 37^\circ = 0.8 \)). The coefficient of static friction is \( \mu_s = 0.5 \) and kinetic friction is \( \mu_k = 0.4 \). Taking \( g = 10\text{ m/s}^2 \), determine whether the block remains at rest or slides down, and find the net force acting on the block along the incline.
View Answer & Solution
Correct answer · Option C
Block slides down; net force = \( 14\text{ N} \)
Explanation
Gravity force along incline \( F_g = m g \sin 37^\circ = 5 \times 10 \times 0.6 = 30\text{ N} \). Maximum static friction \( f_{s,\max} = \mu_s m g \cos 37^\circ = 0.5 \times 50 \times 0.8 = 20\text{ N} \). Since \( F_g > f_{s,\max} \), block slides down. Kinetic friction \( f_k = 0.4 \times 50 \times 0.8 = 16\text{ N} \). Net force \( F_{\text{net}} = 30 - 16 = 14\text{ N} \).
