There are several points I wouldn't subscribe, both in problem's
statement and in the answers I've read.
Both problem and answers insist in saying of "work done on Jerome" and
"work done by Jerone". In correct mechanics terminology, work is done by forces acting on moving points. Not by students or on students.
Another error is in believing that only work done by external forces
matters. But in many cases internal forces too do work and contribute
to energy balance. This is one of such cases.
There is an important difference, often overlooked, about the
rôle of internal forces. They don't count as to centre-of-mass
motion, since their resultant always cancels because of Newton's third law. But internal forces can do work if the body isn't rigid. And human body is not.
Let's examine in greater detail the physical situation. We're said that
Jerome runs upstairs at a constant speed. What we can safely deduce is
that external net force is zero.
Now forget internal forces (as everybody did until this post) and
look for external forces and their work. There are two:
Gravitational force on Jerome's body actually consists of a
multitude of elementary forces, acting on each particle of that body.
It would appear an impossible task to compute their total work, were it
not for a general theorem: if gravitational field is uniform then the
total work of gravitational forces equates that done by total weight
acting on body's centre of mass. So its expression is simply
$-mgh$ - a negative work for an upstairs run.
Upward force applied by each step's tread to Jerome foot. Actually
that force isn't constant but in the average it equals (in opposite
sense) Jerome's weight. What's its work? Note that while a foot is
touching a step its velocity is zero and when it moves it doesn't
touch a step. So work of this force is always zero. We see that total work of external forces is $-mgh$.
Now about the problem's question:
Determine the work done by Jerome in climbing the stair case.
We could translate this question as follows: there are forces applied by
Jerome's body to external world. Compute their total work.
Now the only such forces are those applied by Jerome's feet to step
treads - action/reaction couples together with the ones considered above.
For the same reason (the stairs doesn't move) that work is zero. And
this should be the answer - surely not what teacher expects!
We don't know (OP didn't say) what physics concepts preceded the problem in the course the OP is taking. So maybe I'm going to apply some physical knowledge the OP didn't yet reach. But I find it unavoidable if we want to understand what the teacher expected when the problem was assigned.
To me, the problem implies energy balance.
In mechanics the first approach to energy balance is the energy/work
theorem: variation of kinetic energy of a system equals the net work
done by external and internal forces applied to system's points.
We know variation of kinetic energy is zero, then total work too
must vanish. But we analyzed work done by external forces and found
it's negative. The only way out is to show that there is a positive
work done by internal forces.
Consider what happens when a foot (say the right one) leaves a step,
reaches the next and the whole body is lifted. Firstly, the right leg
is contracted - some work is done but I'll neglect it to simplify
things. Then right foot leans against the next tread. At this point
right gluteal and leg muscles contract, the whole leg extends and the
rest of body is lifted. During this contraction positive work is done
as muscles' ends get closer to each other.
In terms of energy this work is done at the expense of muscles'
chemical energy and this is why Jerome would get tired if his run
lasted much more than 1.32 seconds.
Just for completeness: where does that energy go? @BobD wrote
So gravity takes the work Jerome does in elevating himself and stores
it as his gravitational potential.
IMO that wording is at risk of confusing the OP. Surely there is a
gravitational potential and a gravitational energy. It's not entirely
clear where such energy is residing:
- does it belong to Jerome's body?
- to the system Jerome + Earth?
- to gravitational field?
In a sense all these options are acceptable at different levels of
deepening. At the most elementary level - the one I guess is the most
appropriate in present situation - I'd choose the first answer.
Gravitational forces are conservative and a potential energy can be
defined. Given a fixed gravitational field - in our case the Earth's -
every body possesses a gravitational energy of its own, which only
depends on its mass and on distance of its com from Earth's centre, or -
if you like better - on its height over a reference plane.
When speaking about work two equivalent approaches are allowed:
- to speak of work done by gravitational force
- to think of variation of potential energy.
The former equates minus the latter. Since we had found $-mgh$ for
work, potential energy increases by $mgh$ and this answers our last
question: energy lost by muscles is foundd as an increment of Jerome's
potential energy.
