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06:12
@JohnRennie Hi :-)
Hi :-)
Can we return to yesterday's discussion?
OK ... ?
19 hours ago, by mo-_-
why Wr has different limits of integration from Wgen
That doesn't make sense to me. Both should integrate to infinity.
06:18
Yes but two different results come out
That is, in Wr is then added ∆KE
So Wgen = Wr + ∆KE
But perhaps the dissipated energy in a time t can also be found as W = ∆KE = 1/2(v0² - vf²)
Give me a moment to find the original question and reread it ...
But I don't understand why sometimes Wr is also added ....
@JohnRennie wait i can pick up another question we solved
Jan 12 at 9:58, by mo-_-
they did this using work
@JohnRennie You can see here
I think that must be a mistake. The upper limit of the integral should be ∞
06:27
So as Wr?
But they added kinetic energy to Wr
I don't understand why
3 mins ago, by mo-_-
Jan 12 at 9:58, by mo-_-
they did this using work
Also here they found E_dis using a shortcut with W = ∆KE
And the result is the same
In this question the energy supplied by the generator goes to two places:
1. the KE of the rod
2. heat dissipated in the resistance
Wᵣ is the heat dissipated in the resistance, so Wgen = Wᵣ + KE
OK so far?
Yes
@mo-_- In this question there is no generator to supply power.
That means the energy dissipated as heat must be equal to the decrease in KE because there is no source of extra energy so the total energy must be constant.
06:33
Ok but then the energy dissipated (if there is no generator that supplies power) , can be Calculated as W = ∆KE
And also as ∫P(t) dt
Ah ok :-)
But does the W = ∆KE method always work, when there is no generator? Or are there cases where I have to do the power integral i.e W = ∆KE doesn't work?
Energy is always conserved.
Right
If there is no energy source like a generator or battery then any energy lost as heat must come from the energy (KE or PE) of the sustem.
In that question the only energy of the system was the KE of the loop, so the KE + heat must stay constant i.e. heat = decrease in KE.
06:37
Oh ok clear :-)
But the way to remember this is that the total energy is constant. That's a principle you can apply to any system where there is no extra source of energy.
Ok :-)
Thank you so much, you cleared my doubt!
You're welcome :-)

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