Why isn't Kinetic Energy always equal to Potential Energy?

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Discussion Overview

The discussion centers around the relationship between kinetic energy (K) and potential energy (U), specifically questioning why they are not always equal. Participants explore theoretical implications, mathematical derivations, and the conditions under which these energies are defined, touching on concepts such as the work-energy theorem and conservative forces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a mathematical derivation suggesting K=U, questioning why this equality does not hold universally.
  • Another participant corrects the first by stating that the force is -dU/dx and emphasizes that potential energy is defined up to an arbitrary constant, leading to the work-energy theorem.
  • A different viewpoint suggests that while changes in kinetic and potential energy may be equal in magnitude, this does not imply that the energies themselves are equal, using a tax analogy for clarification.
  • Several participants express confusion regarding the derivation of kinetic energy from momentum, asking for clarification on the integral approach used.
  • One participant notes that the assumption of a conservative force is critical to the derivation and points out that non-conservative forces, like friction, complicate the relationship between K and U.
  • A later reply introduces a more abstract viewpoint, suggesting a different starting point for the relationship between kinetic energy and momentum, which may generalize to special relativity.

Areas of Agreement / Disagreement

Participants do not reach a consensus. There are multiple competing views regarding the relationship between kinetic and potential energy, the assumptions involved, and the implications of non-conservative forces.

Contextual Notes

Participants highlight limitations in the assumptions made about forces being conservative and the definitions of potential energy, which may affect the validity of the derived relationships.

phasacs
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K= ∫mvdv = ∫m dx/dt dv = ∫m dx/dv dv/dt dv = ∫m dv/dt dx = ∫Fdx = U
=> K=U, why isn't this true? If it is, wouldn't that mean that Kinetic Energy is always equal to Potential Energy?
 
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The force is -dU/dx, not dU/dx. Furthermore, the potential is only defined up to an arbitrary constant. If you account for these things, what you get is just the work-energy theorem.
 
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What you found is that the change in KE is equal to the negative of the change in PE. Or the total change is zero. This is "conservation of "mechanical energy".
The fact that changes are equal in magnitude do not grand equality of the actual quantities.

Every time I pay taxes, the amount i give is equal to the amount they receive. But my bank balance is not even close to the one of the internal revenue agency.:)
 
phasacs said:
K= ∫mvdv = ∫m dx/dt dv = ∫m dx/dv dv/dt dv = ∫m dv/dt dx = ∫Fdx = U
=> K=U, why isn't this true? If it is, wouldn't that mean that Kinetic Energy is always equal to Potential Energy?
Kinetic energy is not integral sum of momentum.. Then how u did this?
 
Orodruin said:
The force is -dU/dx, not dU/dx. Furthermore, the potential is only defined up to an arbitrary constant. If you account for these things, what you get is just the work-energy theorem.
How he found kinetic energy in terms of integral sum of momentum?? Can you explain sir?
 
STAR GIRL said:
How he found kinetic energy in terms of integral sum of momentum??
Isn't it the "change" in KE as nasu said earlier?
kinetic energy E=(1/2)*mv2
You can see dE/dv=mv.
So, dE=mv*dv.
 
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cnh1995 said:
Isn't it the "change" in KE as nasu said earlier?
kinetic energy E=(1/2)*mv2
You can see dE/dv=mv.
So, dE=mv*dv.
Oh I see
 
phasacs said:
K= ∫mvdv = ∫m dx/dt dv = ∫m dx/dv dv/dt dv = ∫m dv/dt dx = ∫Fdx = U
=> K=U, why isn't this true? If it is, wouldn't that mean that Kinetic Energy is always equal to Potential Energy?

Note that you have made an assumption that the net force (m dv/dt= F) is a conservative force ( ∫Fdx = [-] U ising @Orodruin 's correction in obtaining a potential energy), which isn't necessarily true... for example, if there is friction that is doing nonzero work.
And as @nasu mentioned, you have really calculated the change-in-K (on the left) and the change-in-[minus]U (on the right), assuming the net force is conservative.

From a more abstract viewpoint [which generalizes to special relativity], it is better to start off with ##\Delta K\equiv\int v\ dp##.
For non-relativistic physics, you will then see that (since v and p are linearly related) ##\int v\ dp=\int p\ dv##.
 
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