# Work Done by Running Up Stairs

This problem is from "the Physics Classroom": During the Powerhouse lab, Jerome runs up the stairs, elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.

a. Determine the work done by Jerome in climbing the stair case.

I am confused by the term "work done" in a). I was thinking that this is not the work done by Jerome but rather the work done just by the vertical component of Jerome's average force, correct? As Jerome is running up the stairs, his average force also has a horizontal component so the work done by Jerome is greater than the answer one would get for a). Secondly, technically isn't the work done by Jerome actually the work being done by the reaction force of the stairs because the reaction force is what is propelling Jerome?

First of all, @Bill Watts is correct on every point. He did it in a very precise way. For that reason I am up voting it. The following simply elaborates on his answer.

The problem you describe given in the Physics Classroom has a lesson associated with it that you have not described. That lesson is described in the following link:

In that lesson they make the following assumptions and statements that are essential to solving this problem. I will summarize them as follows.

1. The ascent of the stairs occurs at constant speed
2. The ascent of the stairs occurs with a constant force equal to the student’s weight.
3. It is assumed that the work which is done by the students contributes to their change in height and not to any change in speed
4. It is assumed that the work done on the student is equal to the work done by the student. The work done on the student equals the students weight times the change in height.

In the context of these assumptions I believe your questions can be addressed.

I am confused by the term "work done" in a). I was thinking that this is not the work done by Jerome but rather the work done just by the vertical component of Jerome's average force, correct?

Not according to the lesson. In connection with assumption 3 above they said giving the students a little head start into the staircase means that whatever force is applied to the step is mostly a downward force which results in an upward reaction force that elevates the student to the next step. So the little head start resolves the issue of how Jerome achieved his constant horizontal velocity. Once a horizontal velocity is achieved, a forward force is not needed to sustain Jerome's horizontal motion (constant horizontal velocity). In physics there is no net force on an object that is moving at constant velocity. Therefore if Jerome's horizontal component of velocity is constant, no physics work is being done in the horizontal direction.

I would point out, however, that Jerome would certainly expend energy moving in the horizontal direction (even if he only moved horizontally at constant velocity) as he would certainly tire. But physics work does not always equal physical effort. Imagine holding a heavy box without moving it anywhere. You will be doing no physics work but you will certainly get tired of holding it after a while.

As Jerome is running up the stairs, his average force also has a horizontal component so the work done by Jerome is greater than the answer one would get for a)

As already pointed out, if he is moving horizontally at constant velocity there is no horizontal component of force. If there were, he would accelerate.

Secondly, technically isn't the work done by Jerome actually the work being done by the reaction force of the stairs because the reaction force is what is propelling Jerome?

From assumption 4, the work done on Jerome is equal to the work done by Jerome. Recall our discussions on this previous post:

Work done by a weight lifter

The work done on Jerome is the work done by the downward force of gravity. The force of gravity (downward) is in the opposite direction to the movement of Jerome (upward) meaning gravity is doing negative work on Jerome. That negative work equals the positive work that Jerome does (force upward and movement upward). So gravity takes the work Jerome does in elevating himself and stores it as his gravitational potential.

Hope this helps.

There is no horizontal force, so if he starts from rest and ends at rest, the horizontal work he did to get moving will be canceled by him coming to a stop. The net work done will be that of opposing gravity to the given height.

If on the other hand, he still moving with the same constant velocity when he gets to the top, the total work will be his increase in kinetic energy plus the work done by raising himself to the given height. The constant velocity will have both horizontal and vertical components, but we only need to know the total velocity to compute the kinetic energy.

The problem doesn't really state whether he stops at the end of his climb.

Update

If we take the problem as stated literally, then we have no starting or stopping, just a constant velocity up the steps. The force required to maintain a constant velocity in the horizontal direction is zero. The force required to maintain a constant velocity in the vertical direction is $$mg$$. The total work is $$mgh$$.

To address some of the comments of no net force in the horizontal direction when there is starting and stopping, I am talking about time average and distance average net force. Starting from rest, accelerating, and then stopping requires:

change in momentum = $$\int Fh(t) \, dt=0$$

where $$Fh$$ is the horizontal force and it time average must be zero. Also with no change in kinetic or potential energy in the horizontal direction we must have:

$$\int Fh(x) \, dx=0$$

so the distance average net force in the horizontal direction must be zero.

• Why doesn't Jerome have a horizontal force? Doesn't "the horizontal work he did to get moving" involve his horizontal force?
– user225790
Jun 2 '19 at 0:14
• And in my answer I pointed out it will be canceled when he stops. Jun 2 '19 at 0:17
• But even if his horizontal work gets canceled, there still is horizontal work done by Jerome and the question asks for work done by Jerome, not net work. So the overall work done by Jerome is greater than the answer to a) because Jerome does work in both the vertical and horizontal directions.
– user225790
Jun 2 '19 at 0:43
• He is doing the net work. Jun 2 '19 at 2:06
• Isn't gravity and his vertical force doing the net work in the vertical direction and in the horizontal direction, it's friction and his horizontal force?
– user225790
Jun 2 '19 at 2:14

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 $$-mgh$$.

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.