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A particle moves with a velocity $(5 \hat{i}-3 \hat{j}+6 \hat{k}) \mathrm{ms}^{-1}$ horizontally under the action of constant force $(10 \hat{i}+10 \hat{j}+20 \hat{k}) \mathrm{N}$. The instantaneous power supplied to the particle is:
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The correct answer is:
$140 \mathrm{~W}$
Instantaneous power is given by
$P=\vec{F} \cdot \vec{V}$
$\begin{aligned} P & =[10 \hat{i}+10 \hat{j}+20 \hat{k}] \cdot[5 \hat{i}-3 \hat{j}+6 \hat{k}] \mathrm{W} \\ & =[50-30+120] \mathrm{W}=140 \mathrm{~W}\end{aligned}$
$P=\vec{F} \cdot \vec{V}$
$\begin{aligned} P & =[10 \hat{i}+10 \hat{j}+20 \hat{k}] \cdot[5 \hat{i}-3 \hat{j}+6 \hat{k}] \mathrm{W} \\ & =[50-30+120] \mathrm{W}=140 \mathrm{~W}\end{aligned}$
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