Work done: kinetic energy or area under F-ds curve?

Starting from $$F=ma = m \frac{dv}{dt} = m \frac{ds}{dt} \frac{dv}{ds} = m v \frac{dv}{ds},$$ leads to work done = integral of F.ds = integral of mvdv = change in KE.

Suppose a variable force is applied to a body. At time $t=0$, $v=0$ and $F=0$. Then the force increases, and the body accelerates and moves forwards. Then a retarding force brings the body back to rest. Is it correct that the body has moved forwards but no net work has been done on the body, as there is no net change in kinetic energy?

• That is correct. Note that, at the point that the object is accelerating, we have some net work done in the direction of motion (that is $F\cdot ds>0$) and then the retarding force causes $F\cdot ds < 0$ by exactly the same amount, hence the change in kinetic energy is zero. – Guillermo Angeris Jan 2 '15 at 7:57