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(D) Given force $\vec{F} = (2t \hat{i} + 3t^2 \hat{j}) \ N$ and mass $m = 1000 \ g = 1 \ kg$.Using Newton's second law,$\vec{F} = m\vec{a}$,we have $\

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$(2t^3 + 3t^5) \ W$

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(D) Given force $\vec{F} = (2t \hat{i} + 3t^2 \hat{j}) \ N$ and mass $m = 1000 \ g = 1 \ kg$.Using Newton's second law,$\vec{F} = m\vec{a}$,we have $\vec{a} = \frac{\vec{F}}{m} = (2t \hat{i} + 3t^2 \hat{j}) \ m/s^2$.Since $\vec{a} = \frac{d\vec{v}}{dt}$,we integrate with respect to time (assuming initial velocity $\vec{v} = 0$ at $t = 0$):$\vec{v} = \int \vec{a} \ dt = \int (2t \hat{i} + 3t^2 \hat{j}) \ dt = t^2 \hat{i} + t^3 \hat{j} \ m/s$.Power $P$ is defined as the dot product of force and velocity: $P = \vec{F} \cdot \vec{v}$.$P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j})$.$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 \ W$.

An object of mass $1000 \ g$ experiences a time-dependent force $\vec{F} = (2t \hat{i} + 3t^2 \hat{j}) \ N$. The power generated by the force at time $t$ is

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