Recent content by Agustin

Sorry. Yes I`m very excited that it works. Doesn`t it seem interesting how much planning can some simple math do? It thrills me

It does? It seems to work just fine for me, when n=1 dA=(l/2)(1+(-1) ^⌊n⌋(2(n-⌊n⌋)-1)=(l/2)(1-(2(1-1)-1))=(l/2)(1-(-1))=(l/2)(2)=l If n=2 dA=(l/2)(1+(-1) ^⌊n⌋(2(n-⌊n⌋)-1)=(l/2)(1+(-1) ^2(2(2-2)-1)=(l/2)(1+(-1))=0

I`ve already checked it, I think the formula is valid.

You`re right. I forgot to check that part, that means my proof of the particular case on page 6 is wrong. I`ll check it and try working out a general formula that works. I greatly appreciate your time.

If I could post on any other forum I would, however I can´t so I have no choice but to post here. Am I to blame for that?

Maybe it is trivial, but I`m not looking for a trivial solution, but the general solution. That equation does not immediately give you the distance asked for, therefore it is a solution yet it isn`t the complete solution. The formula I present on the document immediately gives the distance asked...

That equation does not work for odd numbers. The problem asks the distance it stops relative to the point B. For example if d=7,25l the block will end up at 0.75l away from B, if we plug in those values in the equation you`ve given: d-2l⌊d/2l⌋=7.25l-2l⌊7.25l/2l⌋=7.25l-2l⌊3.625⌋=7.25l-6l=1.25l

I want to know if my reasoning is right. Wether I made mistakes or not.

Homework Statement explained on document attached Homework Equations Energy on a spring and work done by friction The Attempt at a Solution Included on document https://docs.google.com/document/d/1FNrmIkkWzyZJNsbGbq_DYMyMZbpyMcAYKYk9iTbdR-4/edit?usp=sharing

Homework Statement So there is a falling object, you have to take into account the boyant force, the pull of gravity and the drag force A time dependent distance equation is what we're looking for Homework Equations Fd=CdApav2/2 Where Fd is the drag force Cd is the drag coefficient A is the...