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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?

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  • $\begingroup$ 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. $\endgroup$ – Guillermo Angeris Jan 2 '15 at 7:57
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No net work has been done on the body as there is no net change to any type of energy associated with that body. (an object pushed up ramp to rest at a height would fit your scenario, have no net change to KE but positive change to PE and there have nonzero work done to it.)

Another way of looking at it is that the first force did work on the body, and the body then did work against the second force (or the second force did negative work on the body), with net zero work done on the body after all is said and done.

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