Imagine you have a point A and a point B. Does it takes the same amount of joules to travel from A to B no matter how you do it (if in neutral conditions e.g. no wind)? For example, walking, riding a bike, taking the plane (the plane needs gas), climbing a mountain and using a hang-glider (going up the mountain takes energy)
2 Answers
It depends how generous your "neutral conditions" are. If we have no dissipative forces (friction, air resistance etc) as well as no wind, then the only energy needed is equal to $mg\Delta h$ in which $m$ is your mass, $g$ is the gravitational field strength, and $\Delta h$ is the height difference between A and B. So, with the highly unrealistic neglect of resistive forces, the energy requirement is independent of the mode of transport.
In practice, though, dissipative forces of all sorts will be very important and will mean that the energy input will vary hugely between modes of transport.
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$\begingroup$ $mg\Delta h$: only for "small" $h$! $\endgroup$– GertCommented Oct 25, 2020 at 23:28
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1$\begingroup$ @Gert Yes indeed, but from Samuel Fyckes' examples, I reckoned that A and B were pretty close to the Earth's surface. $\endgroup$ Commented Oct 25, 2020 at 23:43
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$\begingroup$ Gravitational forces are conservative even when the field isn’t uniform. $\endgroup$– G. SmithCommented Oct 26, 2020 at 0:37
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$\begingroup$ @G. Smith. Yes, but I think that Gert's point was that the 𝑚𝑔Δℎ formula that I quoted is applicable only if the field is uniform or if $\Delta h$ is small. $\endgroup$ Commented Oct 27, 2020 at 12:35
If your field is conservative only the lenght against the flux or potential matters.
Moving from A to B is hard work. We define work as the force integrated the path. This is the minimum energy needed to move from A to B. Therefore depending on your efficiency you multiply by some number defined as 1-inefficiency.
So no it really depends on how you define your integral and what your field is and what exactly you are integrating, as well as your underlying assumptions