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InterviewSolution
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Prove work energy theorem for a variable force. |
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Answer» Using the relation for kinetic energy, m = Mass of a body u = Initial velocity of the body v = Final velocity of the body a = Acceleration ki = Initial kinetic energy of the body kf = Final kinetic energy of the body The rate of change of kinetic energy is, \(\frac{dk}{dt} = \frac{d}{dt} \Big(\frac{1}{2} mv^2 \Big)\) \(= m \frac{dv}{dt} \times v\) = Fv [\(\because F =m \frac{dv}{dt}\) (Newton's second law)] \(\because \frac{dk}{dt} = F \frac{dx}{dt}\) dk = Fdx On integrating from the initial position (xA) to the final position (xB), we have \(\int \limits ^{k_f}_{k_i} dk = \int \limits ^{x_B}_{x_A} Fdx\) kf - ki = \(\int \limits ^{x_B}_{x_A} Fdx\) = W \(\therefore\)Work done on the body = Increase in K.E. of the body. |
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