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Physics · WORK, ENERGY AND POWER
हिन्दीA body of mass \( 2\text{ kg} \) initially at rest on a horizontal surface is subjected to an applied horizontal force of \( 7\text{ N} \). If the coefficient of kinetic friction is \( \mu_k = 0.1 \) and \( g = 9.8\text{ m s}^{-2} \), the net work done on the body in a time interval of \( 10\text{ s} \) is:
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Correct answer · Option A
\( 635.04\text{ J} \)
Explanation
Frictional force \( f_k = \mu_k mg = 0.1 \times 2 \times 9.8 = 1.96\text{ N} \). Net force \( F_{\text{net}} = 7 - 1.96 = 5.04\text{ N} \). Acceleration \( a = 5.04 / 2 = 2.52\text{ m s}^{-2} \). Displacement \( s = \frac{1}{2} a t^2 = 0.5 \times 2.52 \times 100 = 126\text{ m} \). Net work done \( W_{\text{net}} = F_{\text{net}} \times s = 5.04 \times 126 = 635.04\text{ J} \).
