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(C) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the net force.$\Delta KE = W_{net} = \int P \, dt$G

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$2$

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(C) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the net force.$\Delta KE = W_{net} = \int P \, dt$Given $m = 2 \ kg$,$P = \frac{3t^2}{2}$,and $v(0) = 0$.$K_f - K_i = \int_{0}^{2} P \, dt$$\frac{1}{2}mv^2 - 0 = \int_{0}^{2} \frac{3t^2}{2} \, dt$Substituting $m = 2 \ kg$:$\frac{1}{2}(2)v^2 = \frac{3}{2} \int_{0}^{2} t^2 \, dt$$v^2 = \frac{3}{2} \left[ \frac{t^3}{3} \right]_{0}^{2}$$v^2 = \frac{3}{2} \left( \frac{8}{3} - 0 \right) = 4$$v = 2 \ m/s$.

Power supplied to a particle of mass $2 \ kg$ varies with time as $P = \frac{3t^2}{2} \ W$. Here,$t$ is in $s$. If the velocity of the particle at $t = 0$ is $v = 0$,the velocity of the particle at time $t = 2 \ s$ will be ............. $m/s$.

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