A force F = -K(yhat{i} + xhat{j}) (where K is | vedclass

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(C) The work done is given by $W = \int \vec{F} \cdot d\vec{r}$.Path $1$: From $(0,0)$ to $(a,0)$ along the $x$-axis.Here,$y = 0$ and $dy = 0$. The di

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$-Ka^2$

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(C) The work done is given by $W = \int \vec{F} \cdot d\vec{r}$.Path $1$: From $(0,0)$ to $(a,0)$ along the $x$-axis.Here,$y = 0$ and $dy = 0$. The displacement vector is $d\vec{r} = dx\hat{i}$.The force becomes $\vec{F} = -K(0\hat{i} + x\hat{j}) = -Kx\hat{j}$.Work $W_1 = \int_{0}^{a} (-Kx\hat{j}) \cdot (dx\hat{i}) = 0$ (since $\hat{i} \cdot \hat{j} = 0$).Path $2$: From $(a,0)$ to $(a,a)$ parallel to the $y$-axis.Here,$x = a$ and $dx = 0$. The displacement vector is $d\vec{r} = dy\hat{j}$.The force becomes $\vec{F} = -K(y\hat{i} + a\hat{j})$.Work $W_2 = \int_{0}^{a} (-K(y\hat{i} + a\hat{j})) \cdot (dy\hat{j}) = \int_{0}^{a} -Ka \, dy = -Ka[y]_{0}^{a} = -Ka^2$.Total work $W = W_1 + W_2 = 0 + (-Ka^2) = -Ka^2$.

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