Work Energy Power

JEE Main level, single correct, medium

146 of 183

Question 146 of 183

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

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Answer: (D) 6:136: 13

⟨p⟩=(2i^+3j^)⋅(3i^+6j^)4=6a→=(F→m=12i^+34j^)v→ at t=4sec=(12i^+34j^)×4=(2i^+3j^)Pins =(2i^+3)(2i^+3j^)=13⟨P⟩Pins =613\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}

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