A body of mass 4 kg is placed on a plane at a point P having coordinate $(3,4) \mathrm{m}$. Under the action…

A body of mass 4 kg is placed on a plane at a point P having coordinate $(3,4) \mathrm{m}$. Under the action of force $\overrightarrow{\mathrm{F}}=(2 \hat{i}+3 \hat{j}) \mathrm{N}$, it moves to a new point Q having coordinates $(6,10) \mathrm{m}$ in 4 sec. The average power and instanteous power at the end of 4 sec are in the ratio of :
  1. $13: 6$
  2. $4: 3$
  3. $1: 2$
  4. $6: 13$

Solution

$\begin{aligned} & \langle\mathrm{p}\rangle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \\ & \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \\ & \overrightarrow{\mathrm{v}} \text { at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \\ & P_{\text {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \\ & \frac{\langle\mathrm{P}\rangle}{P_{\text {ins }}}=\frac{6}{13}\end{aligned}$

Asked in: JEE Main 2025 (28 Jan Shift 2)

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