Recent content by davieddy

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...

L (in your formula) is the distance of the center of mass from the pivot, not the L specified in the question.

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...

d = (u+v)t/2 (area under v-t graph or average velocity * time) u=0 PLEASE try harder:)

As I said t = 2d/v You know v and d What's the problem?

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...

Yep!

In part b) you have added the max KE to the max PE. When KE is max, PE is zero and vice versa.