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

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 $y$-axis. What power will be developed by the force at the time $t$?

Accepted Answer

1. **Identify Acceleration:** According to Newton's Second Law, $\vec{F} = m\vec{a}$. Given $m = 1	ext{ kg}$, the acceleration is $\vec{a} = \vec{F}/1 = 2t\hat i + 3t^2\hat j$ [Class 11 Physics, Ch 4, Sec 4.5, Eq 4.5].
2. **Calculate Velocity:** Velocity is the time integral of acceleration. Assuming the body starts from rest ($v=0$ at $t=0$):
   $$ \vec{v} = \int \vec{a} dt = \int (2t\hat i + 3t^2\hat j) dt $$
   $$ \vec{v} = t^2\hat i + t^3\hat j $$
3. **Calculate Power:** Instantaneous power ($P$) is the dot product of force and velocity.
   $$ P = \vec{F} \cdot \vec{v} $$ [Class 11 Physics, Ch 5, Sec 5.10, Eq 5.21]
   $$ P = (2t\hat i + 3t^2\hat j) \cdot (t^2\hat i + t^3\hat j) $$
   $$ P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5	ext{ W} $$

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