Conservation of energy and velocity dependent potential [closed]

Question: Why is energy not conserved when the potential is dependent on velocity? If not, then why not?

Attempt: I know that if the potential is dependent on time, then energy is not conserved. A velocity $$v$$ can also be expressed as $$v=x/t$$, from where an explicit time dependents arises, but I'm not so sure if this is the right way.

edit clarity: The Hamiltonian is not the total energy if $$V=V(\mathbf{q},\mathbf{\dot{q}})$$ but can still be conserved from my understanding. Here I mean the total energy $$E=T+V$$. Is $$E$$ still conserved even when $$V=V(\mathbf{q},\mathbf{\dot{q}})$$? I think I struggle a bit with differentiating in which cases $$H$$ is the total energy and/or conserved $$and$$ $$E=T+V$$ is the total energy (which should always be the case right?) and/or conserved.

• Commented Aug 1 at 17:00
• velocity is $x/t$, but only if it is constant. Commented Aug 1 at 18:40
• You should tell us how you are defining "energy." Do you mean $pv-L$, which is the Hamiltonian energy? Or do you mean $T+U$, which can be called the "total mechanical energy." Those two expressions are not necessarily the same when the potential depends on the velocity, since $L=T-U$ and $p=\frac{\partial L}{\partial v}$...
– hft
Commented Aug 1 at 21:39
• thanks for pointing that out! I hope the edit helps
– Jowo
Commented Aug 2 at 13:02
• o7 is the edit enough for clarity, or should I add something else? Or should I completely rephrase the question?
– Jowo
Commented Aug 4 at 18:37