IDENTIFY and SET UP: The force $\overrightarrow{F}$ is not constant, and in general it is not in the same direction as the displacement. To calculate the work done by $\overrightarrow{F}$. we'll use the general expression for work, Eq.(6.14): $W={\int }_{p_1}^{p_2}\overrightarrow{F}.\overrightarrow{dl}$

where $\overrightarrow{dl}$ is an infinitesimal displacement. We'll calculate the work done on each leg of the square separately, and add the result to find the work done on the round trip. If this round-trip work is zero, force $\overrightarrow{F}$ is conservative and can be represented by a potential energy function.

EXECUTE: On the first leg, from $\left(0,0\right)$ to $\left(L,0\right)$, the force is every where perpendicular to the displacement. So $\overrightarrow{F}.\overrightarrow{dl}=0,$and the work done on the first leg is $W_1=0$ .The force has the same value $\overrightarrow{F}=CL\hat{j}$ everywhere on the second leg, from $\left(L,0\right)$ to $\left(L,L\right)$. The displacement on this leg is in the $+y$-direction, so $\widehat{dl}=dy\hat{j}$ and  $\overrightarrow{F.}\overrightarrow{dl}=CLj.dy\hat{j}=CLdy$

The work done on the second leg is then $W_2={\int }_{(L,0)}^{(L,L)}\overrightarrow{F}.\overrightarrow{dl}={\int }_{Y=0}^{y=L}CL dy=CL{\int }_0^{L}dy=C{L}^2$

On third leg, from $\left(L,L\right)$  to $\left(0,L\right).\overrightarrow{F}$ is again perpendicular to the displacement and so $W_4=0.$ The work done by $\overrightarrow{F}$ on the round trip is therefore $W=W_1+W_2+W_3+W_4=0+C{L}^2+0+0=C{L}^2$

The starting and ending points are the same but the total work done by $\overrightarrow{F}$ is not zero. This is a non conservative force; it cannot be represented by a potential-energy functions.