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: