flaw
Smash Apprentice
In general it is, yes, I didn't think anyone would notice. But for work, the integral is path independent if the forces involved are conservative forces (that is, forces that do not change internal energy or however you want to look at it). Actually, the definition of a conservative force is that the work done by that force is path independent. Look up adiabatic process. So throw any kind of friction in there and the semicircle path and the straight path are different. I don't think the wikipedia makes this distinction for some reason.Hmm, Wikipedia gives the work as
W = \int_C F \dot dx = \int_C F \dot v dt
where W is the work, C is the curve being traversed, F is the vector function describing the force, x the position vector, and v the velocity vector.
It seems to me that the second integral (and probably the first) should always at least be solvable numerically.
Regarding your edit: I thought that in general, that integral was path dependent. Is the fact that integrating along the semi-circle and integrating along the line are equivalent just an artifact of the example you chose?
Those two integrals are the same. If velocity is just dx/dt, then substitude that in in the second equation and you get the first.
As for me saying it's not always possible, I meant more that in the real world this doesn't always work out so nicely. What if you have an arbitrary relationship between the vectors? I've done a lot of computational physics on real life problems so my first idea is to plug the data into a computer and let it do the work for me (no pun intended).
But if you feel that is untoward, I challenge you to solve the one dimensional case where you move a particle from -1 to 1 in a field with constant F = sin(1/x) (not sure if this is possible or not, i'm not a math guy)