Medium MCQ +4 / -1 PYQ · JEE Mains 2024

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

  1. A $\left(3 t^3+6 t^5\right) W$
  2. B $\left(9 t^5+6 t^3\right) W$
  3. C $\left(6 t^4+9 t^5\right) W$
  4. D $\left(9 t^3+6 t^5\right) W$ Correct answer

Solution

<p>$$\begin{aligned} & \vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N \\ & \vec{F}=m \vec{a}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) \\ & \vec{a}=\frac{\vec{F}}{m}=\left(3 t \hat{i}+3 t^2 \hat{j}\right) \\ & \vec{v}=\int_\limits0^t \vec{a} d t=\frac{3 t^2}{2} \hat{i}+t^3 \hat{j} \\ & P=\vec{F} \cdot \vec{v}=\left(9 t^3+6 t^5\right) W \end{aligned}$$</p>

About this question

Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force

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