I "A system tries to minimize total potential energy"

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The discussion centers on the principle of minimizing potential energy in systems, particularly in the context of an object launched from Earth at escape velocity. It questions whether this principle applies to unbound systems, suggesting that it may only be relevant to bound systems that can exchange energy irreversibly. The conversation includes a mathematical example of a harmonic oscillator, illustrating how kinetic and potential energies fluctuate while maintaining constant total energy. It emphasizes that systems do not "choose" to minimize energy, as this anthropomorphism misrepresents their behavior. The conclusion highlights that energy transfer is a fundamental aspect of systems with multiple energy reservoirs.
Swamp Thing
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While reading this thread on Stack Exchange... https://physics.stackexchange.com/q...oes-a-system-try-to-minimize-potential-energy ... a question came to mind : -

Say an object is launched away from Earth at a velocity greater than the escape velocity. This system will not end up with its potential energy less than the initial value. Apparently, therefore, we need to qualify the principle of "tendency towards minimum potential energy" so as to exclude such cases? If so, how would we do that rigorously?
 
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As far as I know it only applies to bound systems that can irreversibly exchange energy with another system
 
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There's energy conservation for a closed system. Usually the kinetic and potential energy both change with time but such that the total energy stays constant. Take the harmonic oscillator as an example:
$$m \ddot{x}=-D x.$$
The general solution is
$$x(t)=x_0 \cos(\omega t -\varphi_0),$$
where the amplitude, ##x_0##, and "phase", ##\varphi_0## are integration constants, and ##\omega=\sqrt{D/m}##.

The kinetic and potential energies are
$$T=\frac{m}{2} \dot{x}^2, \quad V=\frac{D}{2} x^2.$$
As a function of time you get
$$T=\frac{m \omega^2}{2} \sin^2(\omega t -\varphi_0), \quad V=\frac{D}{2} x_0^2 \cos^2(\omega t-\varphi_0).$$
Now ##m \omega^2=D## and thus the total energy
$$E=T+V=\frac{D}{2} x_0^2 [\sin^2(\omega t-\varphi_0) + \cos^2(\omega t-\varphi_0)]=\frac{D}{2} x_0^2=\text{const}.$$
 
You're anthropomorphizing inanimate systems. They hate it when you do that.

If you have a system with two energy reservoirs, of any kind (potential and kinetic is but one example), and all the energy is in one, the only thing the system can do with the energy is move it to the other. There is nothing more to this than "if you're all the way to the left, the only direction you can move is to the right".

Swamp Thing said:
While reading this thread on Stack Exchange..
Is that's confusing you, maybe you should go elsewhere.
 
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Thread 'Why higher speeds need more power if backward force is the same?'
Power = Force v Speed Power of my horse = 104kgx9.81m/s^2 x 0.732m/s = 1HP =746W Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same? I understand that if I increase weight, it is hrader for horse to pull at higher speed because now is backward force increased, but don't understand why is harder to pull at higher speed if weight(backward force)...

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