A particle moves under the influence of a force vec{F} = (2xhat{i} + 3y^2hat{j})text{ N}. The work done by this force in moving the particle from the…

The work done is $W = \int \vec{F} \cdot d\vec{r} = \int_{0}^{2} 2x dx + \int_{0}^{3} 3y^2 dy$. Evaluating these integrals: $\int_{0}^{2} 2x dx = [x^2]_0^2 = 4$, and $\int_{0}^{3} 3y^2 dy = [y^3]_0^3 = 27$. Total work $W = 4 + 27 = 31\text{ J}$? Wait, let's re-evaluate: $[y^3]_0^3 = 27$, wait $4 + 27 = 31$. Let's check options. Ah, let's make it $31\text{ J}$ as option B. Let's recalculate: $W = [x^2]_0^2 + [y^3]_0^3 = 4 + 27 = 31\text{ J}$. Let's change force to $3x^2\hat{i}$ for option check: if $\vec{F} = (3x^2\hat{i} + 2y\hat{j})$, then $W = [x^3]_0^2 + [y^2]_0^3 = 8 + 9 = 17\text{ J}$. Let's keep original: $\int 2x dx = 4$, $\int 3y^2 dy = 27$, sum is $31\text{ J}$. So option B is correct.