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Physics · LAWS OF MOTION
हिन्दीA block of mass \( 5\text{ kg} \) is pulled along a rough horizontal surface by a force \( F = 25\text{ N} \) applied at an angle \( \theta = 37^\circ \) above the horizontal (\( \cos 37^\circ = 0.8, \sin 37^\circ = 0.6 \)). If the coefficient of kinetic friction between the block and floor is \( \mu_k = 0.2 \) and \( g = 10\text{ m/s}^2 \), the acceleration of the block is:
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Correct answer · Option B
2.6 m/s²
Explanation
Normal force \( N = mg - F\sin 37^\circ = 5(10) - 25(0.6) = 35\text{ N} \). Kinetic friction force \( f_k = \mu_k N = 0.2 \times 35 = 7\text{ N} \). Net force along horizontal \( F_{\text{net}} = F\cos 37^\circ - f_k = 20 - 7 = 13\text{ N} \). Thus, acceleration \( a = \frac{13}{5} = 2.6\text{ m/s}^2 \).
