Feel free to PM me anytime!
David
(King of adults)
BTW I'm not sure what miniscule-tim was trying to get at when
he said pretend the battery wasn't there.
I would say pretend there was a battery of E/3.
Then you close the switch and nothing happens.
Now insert a battery with emf...
The "final equation" doesn't need force, work, energy, momentum,
and doesn't even explicitly include acceleraration.
Have a good day.
Bye (before I get barred again for my attitude)
David
PS I thought of something else to say, but mercifully for both of us
I have temporarily...
What is false? Constant acceleration of his c of m during the jumping?
You are making the same assumption when you say work = Fd
and mv = (F-mg)t.
BTW although "F" is the force exerted by him on the floor (and vice versa)
neither are doing any work on each other since the floor isn't...
Yes. Let's follow this through and we will find why Ritwik's first answer is right:
Take sign convention as positive = clockwise round the circuit.
With no battery, a charge q flows in the negative direction
until (CE/2 - q)/(3C/2) = q/C
giving q = CE/5.
Now insert the battery, and a...
Sorry can't read your black boxes.
As you have noted (about 10 times) we need to know the time he spends jumping
(in contact with the floor).
If the acceleration was varying, this time might be anything.
Constant acceleration is both reasonable and simple. (He exerts a constant
force...
Power = work/time
Work = mg(d+h)
Time = 2d/v
mv^2/2 = mgh
You measure d and h with the tape measure, and m with the scales.
As usual in this forum, you assume g is known.
If you read Dr Al's "sticky" post on formulae,
you will see "average velocity = displacement / time"
and for constant acceleration,
"average velocity = (u+v)/2"
It follows that d = (u+v)/2 * t.
I don't know why he didn't include this in his list of constant acceleration
equations...
He crouches to a depth d, accelerates to speed v in time t and then ascends
under gravity to a height h.
mv^2/2=mgh
So we know v.
d=vt/2 so t=2d/v
Work done during jump = mv^2/2 + mgd
Power = work/t
Measure how far you crouch down before you jump.
Assume constant acceleration, and you can calculate the time taken
and acceleration during the actual jumping (straightening the legs).
I endorse this sentiment.
By talking about the potentials, not only does the meaning of "potential difference"
(e.g. VA - VB) become clear, but Kirchoff's "potential drops"
law is rendered redundant (satisfied trivially).
In this problem there are only two different potentials involved...