Power supplied to a particle of mass 2, kg va | vedclass

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Question

From work$-$energy theorem,

$\Delta_{\mathrm{KE}}=\mathrm{W}_{\mathrm{net}}$

or $\quad \mathrm{K}_{\mathrm{f}}-\mathrm{K}_{i}=\int \mathrm{P} \m

Vedclass · Accepted Answer

$2$

Vedclass · Answer

From work$-$energy theorem,

$\Delta_{\mathrm{KE}}=\mathrm{W}_{\mathrm{net}}$

or $\quad \mathrm{K}_{\mathrm{f}}-\mathrm{K}_{i}=\int \mathrm{P} \mathrm{dt}$

or $\quad \frac{1}{2} \mathrm{mv}^{2}=\int_{0}^{2}\left(\frac{3}{2} \mathrm{t}^{2}ight) \mathrm{dt}$

$\mathrm{Or}$            $\mathrm{v}^{2}=\left[\frac{\mathrm{t}^{3}}{2}ight]_{0}^{2}$

$	herefore \quad \mathrm{v}=2 \mathrm{m} / \mathrm{s}$

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

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