A body of mass 2 kg begins to move under the action of a time dependent force given by overrightarrow F = (6 t hat i +6 t 2 hat j ) N . The power developed by the force at the time t is given by:

Physics · Work, Power & Energy

JEE Main 2024 — 31 January, Shift 2 — Question 46

A body of mass 2 kg begins to move under the action of a time dependent force given by $\overrightarrow{\mathrm{F}}=\left(6 \mathrm{t} \hat{\mathrm{i}}+6 \mathrm{t}^{2} \hat{\mathrm{j}}\right) \mathrm{N}$ . The power developed by the force at the time $t$ is given by:

Option A:
$\left(6 t^{4}+9 t^{5}\right) W$
Option B:
$\left(3 \mathrm{t}^{3}+6 \mathrm{t}^{5}\right) \mathrm{W}$
Option C:
$\left(9 t^{5}+6 t^{3}\right) W$
Option D:
$\left(9 t^{3}+6 t^{5}\right) W$
Correct

Answer: D

Step-by-step solution

$\overrightarrow{\mathrm{F}}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) N$

$\overrightarrow{\mathrm{F}}=m \overrightarrow{\mathrm{a}}=\left(6 \mathrm{ti}+6 \mathrm{t}^{2} \hat{\mathrm{j}}\right)$

$\vec{a}=\frac{\vec{F}}{m}=\left(3 t \hat{i}+3 t^{2} \hat{j}\right)$

$\overrightarrow{\mathrm{v}}=\int_{0}^{\mathrm{t}} \overrightarrow{\mathrm{a}} d \mathrm{t}=\frac{3 \mathrm{t}^{2}}{2} \hat{\mathrm{i}}+\mathrm{t}^{3} \hat{\mathrm{j}}$

$P=\vec{F} \cdot \vec{v}=\left(9 t^{3}+6 t^{5}\right) W$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Physics
Chapter: Work, Power & Energy
Topic: Power