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

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Question

$P=F v=\operatorname{mav}$

$P=m\left(\frac{d v}{d t}\right) v=\frac{3 t^{2}}{2}$

$\int_{0}^{v} v d v=\int_{0}^{2} \frac{3}{4} t^{2} a t$

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Vedclass · Accepted Answer

$2$

Vedclass · Answer

$P=F v=\operatorname{mav}$

$P=m\left(\frac{d v}{d t}\right) v=\frac{3 t^{2}}{2}$

$\int_{0}^{v} v d v=\int_{0}^{2} \frac{3}{4} t^{2} a t$

$\left[\frac{v^{2}}{2}\right]_{0}^{v}=\frac{3}{4}\left[\frac{t^{3}}{3}\right]_{0}^{2}$

$\mathrm{v}=2 \mathrm{m} / \mathrm{s}$

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