Work done by a person climbing stairs, who or what does the work? I've seen other questions like this but didn't really see any answers.
When a person climbs stairs, the object is the person. Yet we say the person did work...how so? Doesn't work mean an external force had to be applied over a distance to make an object have a change in KE or PE?
Isn't the work more less from the muscles of the person. The stairs didn't do work. Then really isn't the work internal work done, where the work done is causing and changing their chemical energy to kinetic?
I get work done on a box or more complex systems. But who or what does the work in climbing stairs if it is the person themselves in the system. Why don't physics books explain these type of examples. Thanks in advance
 A: A person is not a rigid body (like a box), so you can't do a simple analysis that treats the person's interaction with the stairs that way.
Better would be to model the person as a mass on top of a piston (or pair of pistons).
As the person climbs, the leg muscles (piston) push the body upward.  The piston is pushing on the moving body, so work is being done.  After the body is raised, the piston is moved to a higher step so the work can be repeated.
The piston also pushes on the stairs, but since the stairs don't move, no work is being exchanged with them.
A: I think you have the right idea.  The situation is equivalent to an engine that delivers power to a car to make it go up a hill.  In that situation, we say the work is done by the engine.  The muscles in a person perform the same function as the engine of a car: they power the body to climb the stairs.  From the work-energy perspective, looking at work as a change in potential energy ($W = -\Delta U$), chemical energy is converted to potential energy (in the $U = mgh$ sense) via the engine or the muscles, so that total energy is conserved but potential energy is not.
Edit: to elaborate on energy conservation perspective, you could model it as chemical energy (muscles activating) converted into kinetic energy (legs start moving) and then to potential energy (person climbs steps). To calculate the work involved, you could treat it as equivalent to a point-mass with an initial upward velocity, decelerating due to gravity until its velocity was zero, which would occur at a distance $h$ . The work would be the force ($−mg$) times the distance ($h$), which is equivalent to the change in potential energy $-\Delta U$. Where would you get the initial velocity? That's the chemical energy converted to kinetic energy.
