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Suppose you have a 2-dimensional Euclidean space where one dimension goes left and right and the other dimension goes up and down where one of the units of length is the meter. Furthermore, suppose there's also a time dimension and one of the units of time is the second, and there's a gravitational field pulling down with acceleration $g$. Furthermore, suppose height is absolute. Furthermore, suppose there's a line in that plane that doesn't change with time whose direction is continuous with position on the line and going in one direction along the line, the component of your velocity going right will always be nonnegative. Now we can define speed along the line for any continuously differential movement on the line. Now let's suppose also that there's a point confined to that line of certain positive mass $m$ which follows Newtonian physics and that point always has just 2 forces acting on it, one going down with magnitude $mg$ and one acting perpendicularly to the part of the track it's on. That makes sense only when the direction of the line is differential let alone continuous at that point. When that's the case, we can show that its acceleration along the track is $-g\sin\theta$ where $\theta$ is the angle of the direction of the track above horizontal so $\theta$ is positive in parts going up and negative in parts going down and is confined between $\frac{\pi}{2}$ and $\frac{-\pi}{2}$.

Let's instead drop the criteria that the slope of the line is differentiable and require only that it's continuous and replace the criteria of the gravitational force and normal force which cannot be defined satisfying those properties with the criterion just simply that its acceleration along the track is $-g\sin\theta$. I define the acceleration along the track as the second derivative with time of how far along the line the point is. I realize that although that's always defined, the actual acceleration of the point, the derivative of its velocity with respect to time, could be undefined at some times. Let's define its potential energy as $mgh$ and it's kinetic energy as $\frac{mv^2}{2}$. Let's define its total mechanical energy as its potential energy plus its kinetic energy. My question is

Can we show that its total mechanical energy is always conserved?

Timothy
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  • I don't understand why this isn't a great question or why my answer isn't a great answer. Would it be possible for somebody to explain why to me? – Timothy May 23 '19 at 02:59
  • Because it seems essentially pointless to me. You're asking for a specific proof to a specific problem, and then answering it yourself. The answer also provides no reasoning for how it comes to the answer. Asking a question and self-answering at the same time is allowed. Asking very specific questions and giving vague equations as answers is unlikely to help anyone else. You've basically asked your own homework and answered it; it doesn't really help anyone else. – JMac May 23 '19 at 12:40
  • @JMac Actually I'm long done university and never got that question as a homework question. I'm guessing the reason I thought this question was probably useful before I asked it is because my questions https://physics.stackexchange.com/q/411470/ and https://physics.stackexchange.com/q/436832/ ended up with a positive score. – Timothy May 24 '19 at 00:47
  • I'm not sure this question is worth closing. I have a real question. Maybe the way it's written really is the best way it can be written. I'm sure a few users have a good memory and if they read this question a bit slower than normal reading speed, they'll remember it and understand it. – Timothy May 25 '19 at 02:18

1 Answers1

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Yes, we can show that as follows.

$\frac{d}{dt}(mgh + \frac{mv^2}{2}) = \frac{d}{dt}(mgh) + \frac{d}{dt}(\frac{mv^2}{2}) = mgv\sin\theta + mv\frac{dv}{dt} = mgv\sin\theta + mv(-g\sin\theta) = mgv\sin\theta - mgv\sin\theta = 0$

Timothy
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