A body of mass $1 ext{ kg}$ begins to move under the action of a time-dependent force $\vec{F} = (2t\hat{i}

Question

A body of mass $1 	ext{ kg}$ begins to move under the action of a time-dependent force $\vec{F} = (2t\hat{i} + 3t^2\hat{j}) 	ext{ N}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along the X and Y axes. What power will be developed by the force at time $t$?

Accepted Answer

1. **Identify Acceleration:** From Newton's Second Law, acceleration $\vec{a} = \frac{\vec{F}}{m}$. Given $m = 1 	ext{ kg}$:
   $$ \vec{a} = \frac{2t\hat{i} + 3t^2\hat{j}}{1} = 2t\hat{i} + 3t^2\hat{j} $$
2. **Determine Velocity:** Velocity is the time integral of acceleration ($\vec{v} = \int \vec{a} dt$). Since the body 'begins to move', initial velocity is zero.
   $$ \vec{v} = \int (2t\hat{i} + 3t^2\hat{j}) dt = \left(\frac{2t^2}{2}ight)\hat{i} + \left(\frac{3t^3}{3}ight)\hat{j} = t^2\hat{i} + t^3\hat{j} $$
   (Refer to NCERT Class 11, Chapter 3 for kinematic integration).
3. **Calculate Power:** Instantaneous power $P$ is the dot product of force and velocity ($P = \vec{F} \cdot \vec{v}$).
   $$ P = (2t\hat{i} + 3t^2\hat{j}) \cdot (t^2\hat{i} + t^3\hat{j}) $$
   $$ P = (2t)(t^2) + (3t^2)(t^3) $$
   $$ P = 2t^3 + 3t^5 	ext{ W} $$
   (Refer to NCERT Class 11, Chapter 6, Section 6.10 for the definition of Power).

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