(D) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta K.E. = K_f - K_i$.Given

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(D) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta K.E. = K_f - K_i$.Given mass $m = 500 \,g = 0.5 \,kg$ and velocity $v = b x^{5/2}$.At $x_i = 0$,$v_i = b(0)^{5/2} = 0$,so $K_i = 0$.At $x_f = 4 \,m$,$v_f = b(4)^{5/2} = 0.25 \times (2^2)^{5/2} = 0.25 \times 2^5 = 0.25 \times 32 = 8 \,m/s$.The work done is $W = \frac{1}{2} m v_f^2 - 0 = \frac{1}{2} \times 0.5 \times (8)^2$.$W = 0.25 \times 64 = 16 \,J$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

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$A$ particle of mass $500 \,g$ is moving in a straight line with velocity $v = b x^{5/2}$. The work done by the net force during its displacement from $x = 0$ to $x = 4 \,m$ is ................... $J$ (Take $b = 0.25 \,m^{-3/2} s^{-1}$)

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$2$
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$4$
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$8$
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$16$

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$A$ particle of mass $500 \,g$ is moving in a straight line with velocity $v = b x^{5/2}$. The work done by the net force during its displacement from $x = 0$ to $x = 4 \,m$ is ................... $J$ (Take $b = 0.25 \,m^{-3/2} s^{-1}$)

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