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(C) The power $P$ is defined as the rate of change of kinetic energy $(K.E.)$ with respect to time: $P = \frac{d(K.E.)}{dt}$.Therefore,the change in k

Vedclass · Accepted Answer

$214$

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(C) The power $P$ is defined as the rate of change of kinetic energy $(K.E.)$ with respect to time: $P = \frac{d(K.E.)}{dt}$.Therefore,the change in kinetic energy $\Delta K.E.$ is given by the integral of power over the given time interval:$\Delta K.E. = \int_{t_1}^{t_2} P \, dt = \int_{2}^{4} (4t^3 - 5t + 2) \, dt$.Evaluating the integral:$\Delta K.E. = [t^4 - \frac{5}{2}t^2 + 2t]_{2}^{4}$.At $t = 4$: $(4)^4 - \frac{5}{2}(4)^2 + 2(4) = 256 - 40 + 8 = 224$.At $t = 2$: $(2)^4 - \frac{5}{2}(2)^2 + 2(2) = 16 - 10 + 4 = 10$.Thus,$\Delta K.E. = 224 - 10 = 214 \, J$.

Power applied to a particle varies with time as $P = (4t^3 - 5t + 2) \, W$,where $t$ is in seconds. Find the change in its $K.E.$ between time $t = 2 \, s$ and $t = 4 \, s$. ............... $J$

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