A particle moves in a force field given by $\vec{F}=-k(y\hat{i}+x\hat{j})$ where $k$ is a constant. If the particle moves from $(0,0)$ to $(a,b)$ along a path, what is the work done by the force?

Solution

1. The force is $\vec{F}=-ky\hat{i}-kx\hat{j}$.
2. Since $\frac{\partial F_x}{\partial y}=-k$ and $\frac{\partial F_y}{\partial x}=-k$, the force is conservative.
3. The potential energy $U$ satisfies $\vec{F}=-\nabla U$, so $U=kxy+C$.
4. Work done $$\begin{array}{rl}W& =-\Delta U\\ & =-(U(a,b)-U(0,0))\\ & =-(kab-0)\\ & =-kab\end{array}$$.
5. Hence the answer is (A).