I've heard the example, that work is path dependent. But whether I climb a mountain directly or in serpentines, in the end it's the same amount of work, with the one difference that it takes me longer to climb the serpentines, but in the end I've done the same amount of work against gravity as if I had climbed directly.
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3$\begingroup$ you are confusing minimum work, which is unique, with actual work but actual work depends on how you do it. If your foot is slipping and you are old and shuffling then you expend more work than is the minimum when climbing a mountain. $\endgroup$– hyportnexCommented Apr 15 at 7:36
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2$\begingroup$ The specific case of you talking about climbing a mountain, that work strictly comes from a conservative force, i.e. it is the gravitational potential energy. That is not path dependent, simply by virtue of having a potential function exist for that type of work. Path dependent work is a much more general situation. You are not considering enough. $\endgroup$– naturallyInconsistentCommented Apr 15 at 7:37
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$\begingroup$ I think there should be examples in almost all textbooks, where you compute the work/heat along different paths in thermodynamic space. That being said, I think this is a good question. $\endgroup$– Tobias FünkeCommented Apr 15 at 7:37
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$\begingroup$ @TobiasFunke you are right but that is not the only thing wrong with the teaching of the subject. $\endgroup$– hyportnexCommented Apr 15 at 7:40
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2$\begingroup$ Congratulations. You have reached the point in your physics education where we leave the frictionless vacuum $\endgroup$– By SymmetryCommented Apr 15 at 7:57
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3 Answers
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But whether I climb a mountain directly or in serpentines, in the end it's the same amount of work, with the one difference that it takes me longer to climb the serpentines, but in the end I've done the same amount of work against gravity as if I had climbed directly.
When we say that "work is not a state function" we have in mind a thermodynamic system, described by the laws of thermodynamic/statistical physics, rather than a system approximated by a material point - the approximation that underlies the application of the Newton laws (as in the example given in the Q.)
The internal energy of a thermodynamic system (which is manifestly a state function) can be changed on a microscopic level, e.g., via collisions between the molecules of the container with a hot wall, and on macroscopic level - e.g., via changing the shape of the container or other macroscopic parameters. The former is called heat, the latter work. Obviously, the same change of the internal energy can be achieved by different combinations of the quantity of heat and work - e.g., via different paths.
This distinction between heat and work is not applicable on the level where we describe an object as a point-like - where all its energy is the mechanical energy of its global motion. However, once we take into account the internal degrees of freedom, the distinction is meaningful - mounting the mountain directly, in an elevator, or by walking in serpentines cost different amounts of internal energy and results in different thermodynamic states of the object (cheerfully taking photos vs. falling on the ground from tiredness.)
Related: Why does holding something up cost energy while no work is being done?
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I've heard the example, that work is path dependent.
Work certainly can be path dependent. But, if the work is done by a conservative force then the work is independent of the path.
But whether I climb a mountain directly or in serpentines... I've done the same amount of work against gravity as if I had climbed directly.
This is because gravity is a conservative force.
In equations, a conservative force $\vec F_c$ can be written as: $$ \vec F_c = -\frac{\partial U}{\partial \vec x}\;, $$ where $U(\vec x)$ is a function of space called a potential energy.
The work done by a conservative force is thus: $$ W_{\vec a\to \vec b} = \int_{\vec a}^{\vec b}d\vec {x}\cdot \vec F_c $$ $$ -\int_{\vec a}^{\vec b}d\vec {x}\cdot \frac{\partial U}{\partial \vec x} = U(\vec a) - U(\vec b)\;, $$ which depends only on the endpoints, not the path. I.e., it is independent of the path taken.
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The "path" in a thermodynamic process is a path in state space, and can involve changing pressure as well as volume. A change in pressure is analogous to changing the gravitational constant, in which case you will do a different amount of work when you climb the mountain. You can even change gravity back to its original strength once you get to the top so that in both cases you will have reached the same final state, however the amount of work done is still different because of the different paths.
