A body of mass m , initially at rest, is acted on by the…

A body of mass $m$, initially at rest, is acted on by the force $\vec F = At\,\hat\imath + Bt^2\,\hat\jmath$ ($A$, $B$ constants). Find the power of this force at time $t$.

Explanation of Method

Power of a force growing in time.

Step 1: Velocity

\[ v_x = \frac{1}{m}\int_0^t At'\,dt' = \frac{At^2}{2m}, \]

\[ v_y = \frac{1}{m}\int_0^t Bt'^2\,dt' = \frac{Bt^3}{3m}. \]

Step 2: Power

\[ P = F_x v_x + F_y v_y = At \cdot \frac{At^2}{2m} + Bt^2 \cdot \frac{Bt^3}{3m}. \]

\[P(t) = \frac{1}{m}\left(\frac{A^2t^3}{2} + \frac{B^2t^5}{3}\right)\]

$P$ $t$ $0$ $1$ $2$ $3$ $\tfrac{A^2t^3}{2m}$ $\tfrac{B^2t^5}{3m}$ $P$

The power is the sum of the powers of the two force components, each acting on its own velocity component (curves in arbitrary units).

Result

\[\boxed{\,P(t) = \frac{1}{m}\left(\frac{A^2t^3}{2} + \frac{B^2t^5}{3}\right)\,}\]

Remark

Check. The original result is correct. As a check, $P = dE_k/dt$ with $E_k = \tfrac{m}{2}(v_x^2 + v_y^2) = \tfrac{A^2t^4}{8m} + \tfrac{B^2t^6}{18m}$.