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Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with $n$ degrees of freedom and Lagrangian of the form: $$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$ where $T$ is quadratic in ...

1

If you call $\chi$ the exergy (availability) then $\chi = U + p_o V - T_o S$ where $p_o, T_o$ are the pressure and temperature of the environment (and are assumed to be constant). To find the maximum amount of useful work that can be extracted form the system it is sufficient to analyze reversible processes only so that $dU=TdS-pdV$ and then the exergy ...

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Carl's first paragraph answers your question (though I disagree with his second paragraph) so this is just an addendum to Carl's answer. It sounds to me as if you are describing an ideal conical pendulum. You're correct that no work is done because the two forces, the string and gravity, act at right angles to the direction of motion so $\vec{F}.\vec{dr}$ ...

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There is force exerted (along the string) thru the point at which the mass's string is attached to whatever is holding it up. If there weren't, yes the ball would fly off. Gravity is a separate force, and will cause the mass to oscillate about the vertical axis. So in fact you won't be able to achive uniform circular motion. Take a look at the toys you ...

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Resolving (Decreasing) energy gradients to lower realized potential states is what drives every process on many levels including evolution. www.intothecool.com Here is a super cool paper that shows the process across universal, biological and socio-economic domains. http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=185965 if ...

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To answer your question, you should first understand when is a system most stable. Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero. We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$ Secondly, it should be able to maintain ...

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A system that is in thermal contact with it's environment will tend towards both a lower energy state and a higher entropy state. Basically, the energy of the system + environment is fixed, but energy will flow between the two until they are in a state of maximum entropy. It might be more informative to ask why systems tend towards increased entropy. What ...

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Well, chemical reactions almost always require heat (energy) to take a place, and almost always release heat upon reaction, so by that logic state when elements is unable to keep reacting is a state with insufficient energy or, in other words, lowest energy state (or we probably should say "lower energy state" then one that required for reactions)

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I'll try to explain with the help of an classical example. Take the situations in the picture above. What you're interested in are the first to cases. The unstable state of equilibrium is such a state that when you slightly displace the ball, it departs from the original position. Being at the top of the hill, it has an excess of potential energy (may it ...

1

Roughly: Becouse $F=-\overrightarrow\nabla U$, with $U$ some potential energy (coud be an effective potential energy). Then, if you aren't in a minimum of potetial, your system isn't in equilibrium. Edit: Can you see that 1. and 2. are stables equilibrium?. I chemistry your effective potential energy is some function called Gibbs free energy.

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