Understanding Spring-Coupled Masses: A Quick Guide

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The discussion clarifies the calculation of potential energy in spring-coupled masses, emphasizing that the total potential energy is determined by the individual stretches of each spring, represented by the second expression provided. Kinetic energy is simply the sum of the kinetic energies of the masses involved, expressed as K = Σ (1/2 mv_i^2). A participant expresses confusion over the editing of the original post, questioning the rationale behind leaving part of the query missing. The original poster explains that they resolved their confusion independently and opted to delete the initial query to avoid wasting others' time. The conversation highlights the importance of clarity in forum communication and the resolution of complex physics concepts.
*best&sweetest*
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[SOLVED] spring-coupled masses

Thanks, Doc Al... I think I understand it now!
 
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*best&sweetest* said:
Is it true that the potential energy U is just the sum of 0.5kx_i^2 with i going from zero to 4, or is it that U = 0.5 k x_1^2 + 0.5 k (x_2 - x_1)^2 + 0.5 k (x_3 - x_2)^2 + 0.5 k x_4^2?
What matters is how much each spring is stretched (or compressed). So the total spring potential energy is given by your second expression. (This is explained on the page that you linked.)

I'm equally confused with the kinetic energy...is it just
K = \Sigma (\frac{1}{2} mv_i^2)?
That's it.
 
*best&sweetest* said:
Thanks, Doc Al... I think I understand it now!

There's a first time for everything. That's a tautology almost. But this type of reply, by editing the very first post, is a first for me in the forum. The answer is left hanging with part of the query missing. I can't but help ask why?
 
Shooting star said:
There's a first time for everything. That's a tautology almost. But this type of reply, by editing the very first post, is a first for me in the forum. The answer is left hanging with part of the query missing. I can't but help ask why?

If you are interested in the question, the entire question is quoted in Doc Al's post, and I managed to understand it by myself, so there is no need for other people to waste their time on it...that's why I deleted it.
 
The more elegant way would be to mark the post as SOLVED.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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