We know when an object is moving in a circular motion on a rough horizontal surface, direction of kinetic friction is constantly. Thus, fto calculate the work doen by friction, we need to use integration. $$W = ∫\vec{F} \cdot \mathrm{d}\vec{s} $$ But since in a circle, the displacement is $0$, why is the work done not $0$?
And if not in a circle, in which case (shape of trajectory) will the $0$ displacement cause the work done to be $0$? Any polygon?
In circular motion, your total displacement $$\oint \mathrm{d}\vec{s}$$ is zero, correct. But for work, you are not integrating just the displacement $\mathrm{d}\vec{s}$, but the quantity $\vec{F}\cdot \mathrm{d}\vec{s}$. At each interval of your integral, this quantity is a scalar, and in the case of friction, always negative. So integrating this quantity is not the same as (or even proportional to) integrating $\mathrm{d}\vec{s}$.
As for your second question, work done by kinetic friction can never be zero, since it always opposes th direction of relative motion between the surfaces. It is always a dissipative force, acting to do negative work on the objects it is in contact with. Hope that helps.
