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Sorry for the relatively long post! Thank you for reading and let me know if there is anything I can clarify/fix.

My textbooks defines Work in the following way:

  1. A measure of the amount of energy transferred between two systems.
  2. Positive work is performed by a system when the force it generates has a component in the direction of the displacement of the point of application"

Let's see if this definition makes sense in a few different examples:


Example 1: A piston

This definition makes a lot of sense in the context of a thermally insulated piston where "System 1" is the molecules inside the piston and "System 2" is the molecules outside the piston. In this example, I will speak from the perspective of the particles in the piston (System 1).

  • When the air molecules of the "outside" system push the piston down, "System 1" does negative work. In other words, it absorbs kinetic energy from the "outside" particles.
  • When the air molecules inside the piston push it up, "System 1" does positive work. In other words, it transfers its kinetic energy to the "outside" particles.

In both these examples, one system "gained" energy while the other "lost" energy. Though the net change is still 0.


Example 2: A moving ball 'A' colliding elastically with a stationary ball 'B' (in space)

During the collision, ball A (System 1) does positive work, thus transferring its kinetic energy to ball B. Ball B (System 2) does negative work, thus absorbing kinetic energy from ball A.

So far, so good for their definition of work.


Example 3: Thrown ball/Falling ball

I am having a hard time applying this definition of work to a thrown ball for two reasons:

  1. There seems to only be one system here: the earth-ball system. What would the other system be?

  2. There doesn't seem to be any energy transfer occurring. Rather, the kinetic energy of the earth-ball system is being transformed into potential energy. When thrown, gravity is doing negative work on the earth-ball system, transforming their energy. Where is the energy transfer?


Using this 3rd example and my logic, one of the following could be an explanation:

  1. My textbook is limiting the definition of work.

  2. The definition is fine, I'm just not understanding something about the ball example.


EDIT: This was my attempt to reconcile example 3 and the textbook's definition of work:

In example 3, we can treat "System 1" as the Earth, and "System 2" as the ball. Also, I am conceptualizing gravity as an invisible spring between the two.

As the ball is moving up, "System 1" (Earth) is pulling on it, doing negative work (displacement of the ball is in the opposite direction of the force on the ball). This negative work absorbs energy from the ball, decelerating it. "System 2" (the ball) is pulling on Earth, doing positive work (the displacement of the earth is in the direction of the force of the ball). This positive work transfers kinetic energy to the Earth, technically accelerating it.

Unfortunately, once the ball starts falling. It seems both Earth and ball are doing positive work (direction of displacement and force are the same for each), thus breaking conservation of energy.

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  • $\begingroup$ did you learn that the work made by a force is force times distance? $\endgroup$ – Wolphram jonny Jul 20 '20 at 18:55
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Why does it work for thermodynamics?

There is a reason why the energy-work relationship in thermodynamics/statistical mechanics cannot be simply translated to classical mechanics. This is because in thermodynamics and kinetic theory of gases, we ignore (at least when starting out) any intermolecular forces between the constituent particles. Thus there are no potentials to be accounted for. The only exchange that takes place is the exchange in kinteic energy, and it makes up fr a comforting and intuitive description of energy transfer.

Why does it not work for classical mechanics?

However, in classical mechanics, you do have forces, which have potentials associated with them. These potentials (or the potential energy corresponding to them) isn't as physical as the kinetic energy in the former case. Thus it is hardto visualize any transfer of energy that takes place in form of change in the potential energy. This is exactly what happens when you throw a ball. The kinetic energy of the earth-ball system reduces as the potential energy of the earth-ball system increases, such that the total energy stays constant. In this case, there is interconversion of energy rather than a transfer from one body to another.

A case in classical mechanics where it works

However, you could also take a case of a moving ball colliding with a stationary one (both placed on horizontal ground). This case showa similar characteristics as the thermodynamic case. The kinetic energy of one of the balls get transferred to another, which is quite similar to what happens between a piston and surrounding gass molecules.

Summary

So at best, I wouod urge you to view both the scenarios in different lights so as to avoid any wrong interpretations/conclusions.

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I am having a hard time translating this analogy to a thrown ball. When the balls moving upward, gravity is transforming the balls kinetic energy to potential. I don't see how the gravitational force, the "outside" system (which I guess in this context can be thought of as a "pool of energy") is transferring anything to the ball.

When the thrown ball moves upward gravity does negative work on the ball. That's because the direction of the gravitational force is opposite to the direction of the displacement of the ball. Negative work means gravity is taking kinetic energy away from the ball and storing it as gravitational potential energy of the ball-earth system.

You can also look at this from the standpoint of the work-energy theorem which states that the net work done on an object equals its change in kinetic energy. In this case the only work done on the ball is the negative work of gravity. Since the work is negative, the change in kinetic energy is also negative so that at the top of the flight of the ball the negative change kinetic energy equals the initial positive kinetic energy given the ball by the thrower for a final kinetic energy of zero. Or

$$W_{net}=-mgh=-\frac{mv_{i}^2}{2}$$

From that equation you can determine the height $h$ reached by the ball given the initial vertical component of velocity, $v_i$.

Hope this helps.

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In Thermodynamics we have several different energies:

  1. internal energy $dU$,
  2. work $\Delta W$, and
  3. heat $\Delta Q$.

If we transfer energy between two systems it makes sense to talk about work done on a system. In contrast, in your ball example, we do not transfer energy between systems. We also do not transfer energy from one type (e.g. internal) to an other type (i.e. heat). Instead, the ball transfers its kinetic energy into potential energy -- both belong to the internal energy. Thus, we have only a single system, and it does not make sense to say that it perform work on itself.

However, if the ball hits the floor and transfers part of its energy into heat (=inelastic collision), the system consists of two parts: The ball and its surrounding. Now, thermodynamical distinction becomes useful again.

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  • $\begingroup$ "...ball transfers its kinetic energy into potential energy..." A ball, all on it's own, cannot have a potential energy. You need to define a system which has that potential energy. $\endgroup$ – user258881 Jul 20 '20 at 19:09

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