I think the matter can be set straight if one considers
total energy conservation which has been formulated as the first law of thermodynamics, $$\Delta E_{int}= W+Q.$$For a multicomponent system, on the left hand side of the equation, ##\Delta E_{int}~## is the change in internal energy of the components and includes changes in their kinetic energy, changes in their potential energy as a result of changes in their spatial configuration when the components are interacting, chemical and biochemical changes, etc.
On the right hand side of the equation ##W## and ##Q## are work and heat that cross the boundary of the system. It's a simple expression of the idea that "to change what's inside something must come in from the outside." I will apply this to the physical situation of a man lifting a book of mass ##m## from floor level to height ##h## above his head at constant speed. I will consider three cases of different systems. In all three ##Q=0.##
Case I: The system is the book.
In the equation ##\Delta U_{int}=0## because ##\Delta K=0## by assumption and there is no change in potential energy because this is a one-component system. Potential energy is defined when the system has at least two components interacting via conservative forces. On the right hand side ##W=W_g+W_{man}## and $$\Delta E_{int}=0=W_g+W_{man}=-mgh+W_{man}\implies W_{man}=mgh.$$This is the same result that one can also obtain with the work-energy theorem.
Case II: The system is the book + Earth
This is the system that the textbook chose. As in case I, the change in kinetic energy of the system is zero. However there is a change in gravitational potential energy because the spatial configuration of the two component system changes as the book is moved farther away from the center of the Earth. The change in gravitational potential energy is, by definition, the negative of the work done by gravity, ##\Delta U_g=+mgh.## Thus, $$\Delta E_{int}=\Delta U_g=W_{man} \implies W_{man}=mgh.$$ This is exactly the same result as in the previous case.
Case II: The system is the book + Earth + man
Here the system is isolated, no work crosses the boundary of the system, therefore the change in internal energy must be zero. As before, ##\Delta K=0## but the potential energy case is a bit more complicated for the three-component system. We have
##\Delta U_{b-e}=mgh =~## the change in potential energy of the book-Earth pair.
##\Delta U_{m-e}=Mg\Delta y_{cm} =~## the change in potential energy of the man-Earth pair as the man's CM is displaced by ##\Delta y_{cm}.##
##\Delta U_{m-b}\approx 0,~## we ignore the change in gravitational potential energy between man and book.
In addition we have the change in internal energy of the man. When he lifts the book, he spends biochemical energy that he needs to replenish by eating. The first law becomes
$$0=\Delta E_{int}=\Delta U_{b-e}+\Delta U_{m-e}+\Delta E_{biochem.}\implies \Delta E_{biochem.}=-mgh-Mg\Delta y_{cm}.$$The expression shows how the biochemical energy expended by the man is split into raising the book and his center of mass.
nasu said:
If you have a projectile at rest relative to the observer and then the projectile explodes into fast moving pieces (relative to the same observer), what external force counts as producing the change in kinetic energy of the system?
In this case, if the projectile initially at rest explodes into ##N## pieces, we have an isolated ##N##-component system. No external force acts on the system and the center of mass remains at rest after the explosion. Assuming that the pieces are non-interacting, the change in internal energy of the system is $$0=\Delta E_{int}=\Delta K+\Delta E_{chem} \implies \Delta E_{chem}=-\frac{1}{2}\sum_{i=1}^Nm_iv_i^2.$$It's a pity that textbooks and instructors don't introduce the first law of thermodynamics in the special case ##Q=0## when they start talking about "energy conservation."
