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Definition of Quasi-static: A quasi-static process is a thermodynamic or mechanical process that occurs very slowly, allowing the system to remain in a state of equilibrium at all times.

While explaining Potential Energy, we take an example by lifting a block of mass 'm' from the ground (H = 0, No kinetic energy and Potential Energy) to H = x, very slowly in "Quasi-static" manner. But according to the definition, if there state of equilibrium, how will the block even be lifted and taken up? After lifting, there can be equilibrium with mg and ext.Force and the block can move with constant velocity. Again at H = x, the block needs to be stopped and there cannot be equilibrium.

And why do we need a quasi-static process to explain Potential Energy in the first place? Why shouldn't the external force just lift it quickly?

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And why do we need a quasi-static process to explain Potential Energy in the first place?

We don’t.

A quasistatic process is not required to explain potential energy. It is only a necessary condition for a process to be considered reversible (i.e. a process that does not generate entropy).

Potential energy depends only on position, not the process that results in that position. In your example the gravitational potential energy will be $mgx$ at height $x$ regardless of how it is lifted.

Hope this helps.

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  • $\begingroup$ You seem to be ignoring the context. There are many possible ways to deal with this potential energy business; you can mathematically postulate that the electric potential, say, is this or that specific function satisfying Poisson's equation and has a certain decaying behaviour at infinity. However, it is much more common for the "gives the force felt by test charge" definition, and this force will differ if the test charge is moving quickly because then it will radiate, it will have a magnetic field component, etc. It is in this sense that quasistatic is also needed. $\endgroup$
    – naturallyInconsistent
    Commented Jul 17 at 3:57
  • $\begingroup$ Since the topic is thermodynamics, and not just basic classical mechanics, I thought it was necessary to be as general as can be and consider potentially dissipative horrible cases. $\endgroup$
    – naturallyInconsistent
    Commented Jul 17 at 3:59
  • $\begingroup$ @naturallyInconsistent To me it just seemed that the OP thinks a quasistatic process is needed for the potential energy to be mgx at H=x. But if you think I’m missing a broader context I would encourage you to post an answer $\endgroup$
    – Bob D
    Commented Jul 17 at 10:35
  • $\begingroup$ Look at the OP's question again. The first statement is clearly from a textbook. The 2nd paragraph is likely what the OP wrote by himself. The textbook definition is clearly trying to teach thermodynamics, with the standard caveat that if some thermal process is not done in the quasistatic limit, then the pressure is not even well-defined. I don't think it is a good idea to ignore what the source of the question. $\endgroup$
    – naturallyInconsistent
    Commented Jul 17 at 11:38
  • $\begingroup$ @naturallyInconsistent “The first statement is clearly from a textbook”. Agree, from a thermodynamics textbook. But the purpose is to define a necessary (but not sufficient) condition for a process to be reversible. “The 2nd paragraph is likely what the OP wrote by himself”. Probably. but I can only guess the motivation. $\endgroup$
    – Bob D
    Commented Jul 17 at 12:27

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