if u compress the spring there will be a force that developes and the potential energy will increase then when released the kinetic energy will increase.
if u compress the spring there will be a force that developes and the potential energy will increase then when released the kinetic energy will increase.
if we know how much the spring is compressed then we can find the potential energy and restoring force. This would be difficult to calculate using kinematics b/c the restoring force will constantly change
So we need a relationship between the acceleration of the spring and the position(amount of compression)
it is possible to solve using f=ma=-(k*delta S)=m(ds/dt)
but its very difficult
if the spring has negligible mass then the alternate method would be...
conservation of momentum(P)
The third method is through the following:
Energy conservation
J is the impulse=delta P
F(in the S direction)=ma(in the S direction)=m(dv/dt)=m(dv/ds)(ds/dt)
which changes variables to position, so..
F=mv(dv/ds)
integrate from initial S to final S, which are velocities of F with respect to position
=integral mv*dv
=delta K
OOOooo0o0o thats decieving, I'm in class don't distract me!:tongue1:Quote:
Originally Posted by phatboislim
dont be posting this educational BULLCRAP on this thread...i'm not in school muthafuccer lol
delta K= - integral k(s-s(@eq))ds = -integral k(delta S)d(delta S) = -k((delta S)^2)/2 evaluated from initial delta s to final
well I am and you'll feel my pain :crazy:Quote:
Originally Posted by phatboislim
weaksos:tongue1:
K(final)+.5k(delta S final)^2=K(initial)+.5k(delta S initial)^2
lol@weaksos
Time for the PRS(personal response System) question of the day!
Question:
I am
(1) Here in body & spirit
(2) Here body but @ Spring break in spirit
(3) I am operating someone else's PRS who is on spring break
5 people chose 3
I chose 1
U(potential energy)=1/2 k(delta S)^2
zero of PE delta S=0, s=s(@eq)
U becomes larger as you travel farther from the equilibrium position
Stick and Slip ex.:
you have a force pulling on a mass connected to another mass via a spring.
As you pull the block to the right, the tension in the spring is increasing, when it reaches the max, then the block will move.
such that f max=mu(static)*N=kx
mu(static) is the coefficient of static friction
:lmfao: :lmfao: :lmfao: :lmfao:
Conservation of energy:
two objects with masses m1 and m2
and there is a spring between them
they collide and compress the spring to a maximum extent and then the particles move away from each other at different velocities
If I know the V1i and V2i (which are in opp directions)
how fast are they moving during this collision?
after the spring reaches eq pos. how fast are the particles moving?
so long as the spring is mass-less then u can calculate the speed before the collision
The max compression can be calculated by first finding the sum of the potential energy
K1+K2+0(the elastic potential engergy before collision)
DO NOT set this = to .5k(delta S)^2
this is wrong based on energy conservation
at the collision the two objects become one and move together, so the potential energy before and after compare as follows:
Pi=Pf
m1(v1)i+m2(v2)i=(m1+m2)V
so...
K1+K2+0=.5k(delta S max)^2+.5m1*V^2+.5m2*V^2
KB with the whoring....jesus...and what the hell...i dont know what u are talking about. good thing u have the 30 sec rule lol.
alrigh /physics lesson on to the library!
im up in this bishh!!!!! :D.......1
3628
^^11,260..... :D my # is higher
whew, at the library now
Im fueling all those stalkers out there with all this location info im giving out :ninja:
i have ur IP addy.... :ninja:
alright checking notebook for notebook check @ noon
hook that up to your GPS and stalk like a pro!Quote:
Originally Posted by TRYMY4.0
*inserts flash drive*
8gigs! Fry's FTMFW!
*loading autocad*
It has to "re-install" every time you logon to a school computer which is teh ghey
*logs into Tsquare*
nm I have to check my email for my corrected DWFs, *switches over to email*