Answer
a) It has been proved that a constant force field does zero work on a particle that moves once uniformly around the circle $x^2
+y^2=1$
b) Yes
Work Step by Step
(a) Let us consider that $F(x,y)=A(x,y) i+B(x,y) j$
Since, $W=\int_C F\cdot ds=\int_C A dx +\int_C B dy$
or, $=\int_0^{2 \pi} A [-\sin t dt]+\int_0^{2 \pi} B [\cos t dt]=p(0)+q(0)=0$
This shows that a constant force field does zero work on a particle that moves once uniformly around the circle $x^2
+y^2=1$
b)Let us consider that $F(x,y)=A(x,y) i+B(x,y) j$
Since, $W=\int_C F\cdot ds=\int_C A dx +\int_C B dy=\int_0^{2 \pi} [k\cos t dt][-\sin t] dt+\int_0^{2 \pi} k \sin t=k \int_0^{2 \pi} -\cos t \sin t +\sin t \cos t dt=0$
Yes, for a force field the work done is zero and $F(x)=kx$.
