Why is the derivative of kinetic energy (0.5mv2) wrt velocity equal to momentum (mv)? Does this have any physical or mathematical significance?
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Derivative of kinetic energy. watch
- Thread Starter
- 31-10-2015 16:16
- 31-10-2015 16:52
This is really just two changes of variable; from s to p and p to v.
But this is entirely speculative so I welcome any corrections.Last edited by 16Characters....; 31-10-2015 at 17:11.
- 31-10-2015 17:01
I finally got around to asking one of the physics teachers at my college and she basically said it's just a coincidence.
Aside from what you'd get from the integration/differentiation process, I don't think there's any other intuitive understanding to be had, unfortunately
- 31-10-2015 17:23
In actuality relativistic effects feature and you get a somewhat messier form for the kinetic energy.
Mathematically, we can see this by considering the definition of work and force.
In the case where we accept that and we compute in one dimension, with a the change in kinetic energy of a particle of fixed mass driven by a force , with initial velocity we see that:
( using the fact that we started at rest.)
Hence by the Fundamental Theorem of Calculus.
EDIT: Lol, this is what happens if you take a shower before typing up stuff - you end up spouting the same stuff as everyone else.Last edited by joostan; 31-10-2015 at 17:27.
- Thread Starter
- 31-10-2015 17:33
Thanks for the clear up guys, much appreciated
- 31-10-2015 19:19
Using a suvat equation not containing time, v2=u2+2as, and rearranging to s we get: s=(v2-u2)/2a
And from newtons second law: F=ma
Substituting for s and F we get:
Assuming the initial velocity was zero, we arrive at:
Edit: Mathsandphysftw: just realised that you are not to derive K.E formula, please ignore meLast edited by Absent Agent; 31-10-2015 at 19:45.