Conservation of kinetic energy and external forces In Goldstein's "Classical Mechanics", at page 360  below eq. (8.84) it is stated that:

"If, further, there are no external forces on the system (monogenic and holonomic), ..., then $T$ (kinetic energy) is conserved..." 

however I do not think that this is completely correct. Indeed, if one defines external forces as those forces that are exerted by particles in the environment on particles in the system, then we are missing the contribution of internal forces whose work is not necessarily zero (think of a two mass and a spring system). Therefore the kinetic energy is not conserved in general. The statement may be rectified if "external" is replaced by "active" (meaning all those forces that are not constraints' reactions) but then we would also need to assume the constraints to be smooth.
I would like to know whether this argument is correct and, in that case, what does Goldstein mean with his statement?

As always, any answer or comment is much appreciated and let me know if I can explain myself clearer!
 A: I believe the quoted statement is incorrect. If there is no net external force  on the system, and if there are no internal dissipative forces (e.g., friction), then total mechanical energy, KE + PE, is conserved, not just KE.
Hope this helps.
A: It's not hard to prove that the time derivative of the kinetic energy of a system equals the total power of the forces (from both internal and external forces), namely
$\dfrac{d K}{dt} = P^{tot} = P^{int} + P^{ext}$.
Proof of the theorem of the kinetic energy. This readily comes from the balance of the momentum. As an example, for a system of mass points, the dynamical equation of the $i^{th}$ point reads
$m_i \ddot{\mathbf{r}}_i = \mathbf{F}_i^{tot} = \mathbf{F}^{e}_i + \mathbf{F}^{int}_i$
that multiplied with the scalar product by the velocity $\mathbf{v}_i = \dot{\mathbf{r}}_i$ gives
$P^{tot}_i = \mathbf{v}_i \cdot \mathbf{F}^{tot} = \mathbf{v}_i \cdot (m_i \dot{\mathbf{v}}_i ) = \dfrac{d}{dt}\left( \dfrac{1}{2} m_i \mathbf{v}_i^2 \right) = \dot{K}_i $
and summing over all the masses we get the total power of the forces acting on the system $P^{tot} = \sum_i P_i^{tot}$, and the kinetic energy equal of the overall system $K = \sum_i K_i$, to finally find
$P^{tot} = \dfrac{d K}{dt}$.
