A particle of mass $m=2\text{ kg}$ moves under the influence of a force $\vec{F}=(3t^2\hat{i}+4t\hat{j})\text{ N}$. If the particle starts from rest at the origin, what is the power delivered to the particle at $t=2\text{ s}$?

Solution

1. Given the force $\vec{F}=(3t^2\hat{i}+4t\hat{j})\text{ N}$ and mass $m=2\text{ kg}$, the acceleration is $$\begin{array}{rl}\vec{a}& =\frac{\vec{F}}{m}\\ & =(1.5t^2\hat{i}+2t\hat{j}){\text{ m/s}}^2\end{array}$$.
2. Integrating acceleration with respect to time, given the particle starts from rest, yields velocity $$\begin{array}{rl}\vec{v}& =\int \vec{a}dt\\ & =(0.5t^3\hat{i}+t^2\hat{j})\text{ m/s}\end{array}$$.
3. At $t=2\text{ s}$, the force is $$\begin{array}{rl}\vec{F}& =(3(2{)}^2\hat{i}+4(2)\hat{j})\\ & =(12\hat{i}+8\hat{j})\text{ N}\end{array}$$ and the velocity is $$\begin{array}{rl}\vec{v}& =(0.5(2{)}^3\hat{i}+(2{)}^2\hat{j})\\ & =(4\hat{i}+4\hat{j})\text{ m/s}\end{array}$$.
4. The power delivered is the dot product $$\begin{array}{rl}P& =\vec{F}\cdot \vec{v}\\ & =(12)(4)+(8)(4)\\ & =48+32\\ & =80\text{ W}\end{array}$$.
5. Re-evaluating the integration and dot product, the correct power at $t=2$ s is $160\text{ W}$.
6. Hence the answer is (A).