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I think you're missing something from your question? For instance, if it is displaced 10 meters from the left to the right over a period of ten seconds, but then moved 3 meters back from right to left for those last 3 seconds, then the work done would only be the force used to move it the 7 meters difference... Work only depends on the total distance, not the path taken.
@spatcher, work is path-dependent. When you walk in the city, you would always want to take shortcuts because then the path is shorter, and you'll not have to work much. But if you take a longer path, you'll have to work more, and you might even get tired. Clear?
Ummm... http://en.wikipedia.org/wiki/Conservative_force "A conservative force is a force with the property that the work done in moving a particle between two points is independent of the path taken" We're talking about physics work here, not the english translation of the word work. The OP mentioned that the object goes to and fro, and doesn't say anything about friction, so the work would ultimately be calculated by just the ultimate distance travelled... so if it goes a positive distance (in one direction) but then goes a negative distance (in the opposite direction) a short distance, adding them together a large positive distance with a small negative distance gives you a smaller positive distance. So you'd subtract the negative work from the positive work, and you'd end up with the same value as having just calculated the work from the ultimate distance you travelled.......
Oh yeah, work is path-independent in the case of conservative forces! I forgot that. Thanks for clearing this up, you're correct.