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Mathematically I understood the reason why, for a conservative force, $\vec{F}=-\vec{\nabla}U$.

Physically this means that a conservative force is always oriented in the direction of greatest decrease of potential energy.

What I do miss is the physical reason why a conservative force do so (i.e. tries always to reduce the potential energy). I tried to look at it in terms of the work done by the force but I could not really find a good physical explanation for that.

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I think the question is posed from the wrong side...

Since you ask for an intuitive reason, it's probably rather the force that is the basic concept, and the potential is derived (hehe, derived by an integral, this sounds contradictory :)) from it (if the force is conservative, i.e. there is a unique value of the integral between two points, independent of the path).

So the observation is rather: the potential energy decreases fastest in the direction of the force; which now is obvious, since for a given magnitude of $\mathrm ds$ the work (scalar product) $F\cdot\mathrm ds$ is maximal in the direction where $F$ and $\mathrm ds$ are parallel. Does that address your question?

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  • $\begingroup$ Thanks for the answer, probably that's a easier way to think about it, nevertheless also in this way I do not see why it is obvious that the energy decreases fastest in the direction of force.. I still don't get the physical reason of that.. $\endgroup$ – Sørën Apr 4 '16 at 10:54
  • $\begingroup$ well, it is not a physical reason, it is rather computational, the work (scalar product) $F\cdot\mathrm ds$ is maximal in the direction where the two factors are parallel (... I edited the answer) $\endgroup$ – Ilja Apr 4 '16 at 10:58

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