JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 21)
A body of mass 1kg begins to move under the action of a time dependent force $$\overrightarrow F = \left( {t\widehat i + 3{t^2}\,\widehat j} \right)$$ N, where $$\widehat i$$ and $$\widehat j$$ are the unit vectors along $$x$$ and $$y$$ axis. The power developed by above force, at the time t = 2s, will be ____________ W.
Answer
100
Explanation
$$
\begin{aligned}
& \overrightarrow{\mathrm{F}}=\mathrm{t\hat{i}}+3 \mathrm{t}^2 \hat{\mathrm{j}} \\\\
& \frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=\mathrm{t\hat{i}}+3 \mathrm{t}^2 \hat{\mathrm{j}} \\\\
& \mathrm{m}=1 \mathrm{~kg}, \int_0^{\hat{v}} \mathrm{dv}=\int_0^{\mathrm{t}} \mathrm{tdt} \hat{\mathrm{i}}+\int_0^{\mathrm{t}} 3 \mathrm{t}^2 \mathrm{dt} \hat{\mathrm{j}} \\\\
& \overrightarrow{\mathrm{v}}=\frac{\mathrm{t}^2}{2} \hat{\mathrm{i}}+\mathrm{t}^3 \hat{\mathrm{j}} \\\\
& \text { Power }=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{V}}=\frac{\mathrm{t}^3}{2}+3 \mathrm{t}^5 \\\\
& \text { At } \mathrm{t}=2, \text { power }=\frac{8}{2}+3 \times 32 \\\\
& =100
\end{aligned}
$$
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