When examining a particle in a conservative force field, what operation must be taken to find the work done on the particle as it travels from point A to point B?
The integral of the Force in the direction of motion, \[ Work = \int_A^B F . d\ell \]
Alright, thanks! I thought it was the line integral of the force along the path (in this case a straight line because it's a conservative field), but I wanted to make sure I didn't have to find f such that [f = (GRAD)F\] and integrate THAT.
Reducing it to one of the most simply examples, think about picking up a 1 kg book and putting in on a table 1 meter off the floor. The work is just the displacement in the direction of the acting force--gravity--in the direction of that force--vertical. So the work is \[ W = F \ell = mg \ell = (1 kg)(9.8 ms^{-2})(1 m) = 9.8 kg.m^2.s^{-2} = 9.8 J \] As gravity is an conservative force, if I pick up the book, take it downstairs, show it to my brother, then bring it back up, throw it in the air, slid it along the floor to amuse the dog. Then pick it up, throw it in a summersault parabolic motion so it finally lands on the table, the total work is still only \( mg \ell \).
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