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QJEEM23Ph6218: JEE Main 2023 Physics - Class 11 JEEM23Ph6218 Solution: A body of mass 1kg begins to move under the action of a time…

A body of mass 1kg begins to move under the action of a time dependent force F→=(ti^+3t2 j^)\overrightarrow F = \left( {t\widehat i + 3{t^2}\,\widehat j} \right) N, where i^\widehat i and j^\widehat j are the unit vectors along xx and yy axis. The power developed by above force, at the time t = 2s, will be ____________ W.

Step-by-Step Verified Solution

amp;F→=ti^+3t2j^amp;mdv→dt=ti^+3t2j^amp;m=1 kg,∫0v^dv=∫0ttdti^+∫0t3t2dtj^amp;v→=t22i^+t3j^amp; Power =F→⋅V→=t32+3t5amp; At t=2, power =82+3×32amp;=100\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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