lightarrow said:
In this case there are no problems, of course. The problem is I that my system is not that, but spring+point mass (or spring+rigid ball).
My purpose is not to solve a specific problem of physics, I know how to do it, but to understand if it's possible to generalize the wok-kinetic energy theorem to non rigid systems as in this case.
Of course, as I wrote in post number 10, the mathematical theorem applies to any system of point-particles. Go back to post 10 and read the conditions of applicability of the mathematical theorem clearly stated there.
In this paper for example:
http://www.google.com/url?q=http://...sntz=1&usg=AFQjCNH3Ccbgxw5HbrGwHwoLE7JEt-H4lg
they propose to generalize the theorem (including dissipative effects) writing:
Wext = Δ(E
k) + ΔU
where U is the internal energy. Considering also the heat Q, they propose to generalize the first principle of thermodynamics writing:
Wext + Q = Δ(E
k) + ΔU.
--
lightarrow
All this is a consequence of the Theorem a wrote in post 10. If the internal forces of the system of point particles are conservative, then you can mathematically prove that:
\Delta K = \sum_i(\frac{1}{2}m_iv^2_i(t_1)) - \sum_i(\frac{1}{2}m_iv^2_i(t_0)) =\sum_i(\int_{t_0}^{t_1}\vec{F_i}(t)\cdot\vec{v_i}(t)dt) = \sum_i(\int_{t_0}^{t_1}(\vec{F_i}_{ext}(t) + \vec{F_i}_{int}(t))\cdot\vec{v_i}(t)dt ) =
= \sum_i(\int_{t_0}^{t_1}\vec{F_i}_{ext}(t)\cdot\vec{v_i}(t)dt) + \sum_i(\int_{t_0}^{t_1}\vec{F_i}_{int}(t))\cdot\vec{v_i}(t)dt) = W_{ext} + \sum_i(U_i(t_0) - U_i(t_1)) = W_{ext} - (\sum_i U_i(t_1) - \sum_i U_i(t_0)) =
= W_{ext} - \Delta U
So obviously:
W_{ext} = \Delta K + \Delta U
Again, if you start with the mathematical theorem from my post nº 10 and you suppose that some of the internal forces are conservative, but some other internal forces are not conservative (i.e. you can split F_int = F_int_cons + F_int_non_cons) then:
\Delta K = \sum_i(\frac{1}{2}m_iv^2_i(t_1)) - \sum_i(\frac{1}{2}m_iv^2_i(t_0)) =\sum_i(\int_{t_0}^{t_1}\vec{F_i}(t)\cdot\vec{v_i}(t)dt) = \sum_i(\int_{t_0}^{t_1}(\vec{F_i}_{ext}(t) + \vec{F_i}_{int}(t))\cdot\vec{v_i}(t)dt ) =
=\sum_i(\int_{t_0}^{t_1}(\vec{F_i}_{ext}(t) + \vec{F_i}_{int cons}(t) + \vec{F_i}_{int non cons}(t))\cdot\vec{v_i}(t)dt ) =
= \sum_i(\int_{t_0}^{t_1}\vec{F_i}_{ext}(t)\cdot\vec{v_i}(t)dt) + \sum_i(\int_{t_0}^{t_1}\vec{F_i}_{intcons}(t))\cdot\vec{v_i}(t)dt) +<br />
\sum_i(\int_{t_0}^{t_1}\vec{F_i}_{intnoncons}(t))\cdot\vec{v_i}(t)dt) =
W_{ext} + \sum_i(U_i(t_0) - U_i(t_1)) + Q = W_{ext} - (\sum_i U_i(t_1) - \sum_i U_i(t_0)) + Q = W_{ext} - \Delta U + Q
So
W_{ext} + Q = \Delta K + \Delta U
It is simply puting names to some of the terms of the general theorem of post nº 10.
Also, is F_i_ext can split into F_i_ext_cons + F_i_ext_noncons, doing exactly the same, you'll get to:
\Delta K = W_{ext non cons} - \Delta U_{ext} - \Delta U_{int} + Q
or
W_{ext non cons} + Q = \Delta K + \Delta (U_{int} + U_{ext})
So if all forces are conservative, then:
0=\Delta K + \Delta (U_{int} + U_{ext})
or
K + U_{int} + U_{ext} = constant
If there are no external forces and all internal forces are conservative:
0=\Delta K + \Delta U_{int}
or
K + U_{int} = constant
You can get some more particular cases. All them are particular cases from the general mathematical theorem stated in post nº 10 ( it is simply splitting some terms and puting some names to some line integrals).
) for W = work done on the system by the external forces. I have edited the post.
