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I Work: Isolated & Non-Isolated Systems Difference

  1. Dec 17, 2016 #1
    Okay, so I've reached the Work Chapters in my textbook, and I've noticed some contradictions, especially in what consists (and what doesn't) an Isolated System, plus the external and/or applied forces.

    For example, in one of the "Speed Questions" it categorizes a single cube as a non-isolated system, the surface as a non-isolated system, and the cube/surface (there is friction) as an isolated system. Later, in an exercise, it describes a cube/surface system with friction as a non-isolated system.

    The problem is, it's got a ton of formulas that he constructs, reconstructs, renames and whatnot, and it's confused the hell out of me. From what I gathered, an isolated system is something that each force exists within (eg a cube, the surface and the earth), and a non-isolated one is something that each (or some) force(s) is/are external (eg just a cube, where the force that the earth exerts onto it is external).

    Can anybody plainly explain to me the nature of those systems, external/applied forces, and how work fits in all this? I mean, I've got about 50 or so equations and formulas at this point, with most of them lumped together and reconstructed in every page. I've memorized most of the formulas and mostly know how to use them in the various exercises, but I'm kinda lost as to why I'm using them. Things were pretty clear fro my High School studies, and I never had any particular trouble with Work and the like, but this book (Physics for Engineers and Scientists, 8th Edition)has confused me a bit.

    I'd really appreciate it if someone could go over the basics briefly.
  2. jcsd
  3. Dec 17, 2016 #2


    Staff: Mentor

    Yes, this is exactly the difference.

    By Serway?

    So work is a transfer of energy. If the forces are internal then work just moves energy around within the system. If a force is external, then energy can leave or enter the system via the external force.
  4. Dec 18, 2016 #3
    Well, that's good, at least I've got that down.


    Okay, yeah, I get that. But, for example, in some exercises I see that he sets two different "places" where Ug = 0, one for each object. So, for example, let's say there are two cubes. One is stationed at a reclining surface, and the other is on top of a spring, that is stationed vertically.Both are connected with a weightless rope. We pull the first cube by h, and let him go (Vi = 0). So, when it's time to do the exercise, he says that Ug = 0 in two ocasions: One, when the cube on the reclining surface is dragged/pulled back by h, and two, when the second cube is back at its original place (on top of the spring, which is unstreched).

    Can we do that? From high school, I knew that you could just pick one place where Ug was 0. I'd never seen a problem where you could set two places.
  5. Dec 18, 2016 #4


    Staff: Mentor

    All that matters is differences in potential energy. If you want to set Ug equal to 0 for both at the same time then that is fine too. You will just get a constant on both sides that cancels out. The author is just recognizing that and setting that canceling term to 0 in advance.
  6. Dec 18, 2016 #5
    So, for each object, I'm free to set Ug zero as I see fit.
  7. Dec 18, 2016 #6


    Staff: Mentor

    If you ever think that you need to set Ug at some place then don't hesitate to do so. If you were free to set it somewhere else then it will drop out automatically. Personally, I would have set them to the same to be safe. I would have carried an extra term in my intermediate calculations, but I would rather do that than confuse myself
  8. Dec 18, 2016 #7
    Okay, thanks for the info! These last few chapters are kinda tricky, but as I progress, through the exercises, things become more clear. Now if only I had more time...
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