How Much Work Is Involved in Climbing Stairs for a 60kg Person?

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In climbing stairs that are 10.0 m high, a person weighing 60.0 kg does 5880 J of work, calculated using the formula W = mgh. The discussion highlights that while vertical work is clear, the horizontal component raises questions about whether work is done in that direction. It is noted that since the force exerted horizontally does not result in displacement, no work is done in the horizontal direction. Factors such as the angle of the stairs and riser heights affect the effort required, but the problem primarily focuses on vertical elevation change. Ultimately, the consensus is that in this specific context, only vertical work is considered relevant.
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Homework Statement


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In climbing stairs that are 10.0 m high, how much work does a person weighing 60.0 kg do?

Homework Equations


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W = Fd

The Attempt at a Solution


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I know that when moving upwards, and when there is no change of kinetic energy from start to finish, the work done is W = mgh. In this case, the answer to the problem would be 5880 J. However, I am not sure this is correct, for it only takes into account the vertical direction. Doesn't the person exert a force through a distance (and thus do work) when he or she is moving in the horizontal direction between each step on the stairs?
 
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the fancy way to write Work is ##W=\int F \cdot d\vec{r} ##

So, my understanding is,
The force is all in the vertical direction due to gravity (there's no mention of any frictional effects on the stairs in the problem), so when the person is walking in the horizontal direction, the angle between the displacement ## d\vec{r} ## and the Force is 90 degrees. The dot product between the two is therefore zero, and no work is being done
 
deedsy said:
the fancy way to write Work is ##W=\int F \cdot d\vec{r} ##

So, my understanding is,
The force is all in the vertical direction due to gravity (there's no mention of any frictional effects on the stairs in the problem), so when the person is walking in the horizontal direction, the angle between the displacement ## d\vec{r} ## and the Force is 90 degrees. The dot product between the two is therefore zero, and no work is being done

But to move in the horizontal direction, doesn't there have to be some horizontal component of force? The person is pushing down on the stair which propels him forward, in the horizontal direction, and at the same time he is pushing up, which propels him vertically. So doesn't this mean that there is work being done by the person in both vertical and horizontal directions?
 
there will be a normal force out of the stair as you walk on them. However, since your foot is situated, no frictional force is moving through any distance, so no Work is being done there.
 
Last edited:
deedsy said:
there will be a normal force into the stair as you walk on them. However, since your foot is situated, this force isn't moving through any distance, so no Work is being done there.

So do you mean to say that in walking in a straight line on a flat plane, you do no work in the physics sense?
 
I don't believe so, as long as you aren't skidding/shuffling while you walk (no frictional forces over a distance)
 
You'd also have to assume there is no air resistance
 
But like you said before, when you are walking you are propelling yourself forward; this wouldn't be possible without friction. But since your foot is not slipping while you push (no displacement), the static friction is doing no work.
 
In any event, the problem statement doesn't furnish any information about the angle of the stairs or anything else which would allow for the evaluation of any work performed except for the change in elevation.
 
  • #10
SteamKing said:
In any event, the problem statement doesn't furnish any information about the angle of the stairs or anything else which would allow for the evaluation of any work performed except for the change in elevation.

If the problem did include the angle that the stairs make with the horizontal, how would this contribute to figuring out if the person is doing work in the horizontal direction? At any rate, doesn't the person, after lifting their leg, have to move that leg forward with their muscles in order to get to the next step? Why wouldn't this be considered doing work in the horizontal direction?
 
  • #11
Mr Davis 97 said:
If the problem did include the angle that the stairs make with the horizontal, how would this contribute to figuring out if the person is doing work in the horizontal direction? At any rate, doesn't the person, after lifting their leg, have to move that leg forward with their muscles in order to get to the next step? Why wouldn't this be considered doing work in the horizontal direction?
Well, a stairway which makes a 60° angle with the horizontal has less horizontal distance to travel than one which makes only a 30° angle.

Also, the riser heights (the vertical distance between steps) will indicate the amount of effort it takes for each step. It's easier to climb a stairway with 9" riser heights than one with 18" riser heights.

In short, all stairways are not created equal.
 
  • #12
SteamKing said:
Well, a stairway which makes a 60° angle with the horizontal has less horizontal distance to travel than one which makes only a 30° angle.

Also, the riser heights (the vertical distance between steps) will indicate the amount of effort it takes for each step. It's easier to climb a stairway with 9" riser heights than one with 18" riser heights.

In short, all stairways are not created equal.

I still don't understand why there is no work in the horizontal direction. Isn't the displacement vector along the hypotenuse of the staircase? Not vertical?
 
  • #13
Mr Davis 97 said:
I still don't understand why there is no work in the horizontal direction. Isn't the displacement vector along the hypotenuse of the staircase? Not vertical?
Let's simplify it by setting the vertical distance to zero, so we only have horizontal motion.
What confuses people is that everyday experience involves starting from rest and reaching some reasonable speed. That does require work. But in the context of the question, thre is no minimum speed requirement, and frictional losses are to be ignored. This means that we could do an infinitesimal amount of work to reach some very small speed then just wait long enough to cover the desired distance. There is no minimum amount of work required, so effectively it is zero.
Another way to look at it is that you may invest a small amount of work to get moving, but, in principle, that amount of work can be recouped from KE at the end of the motion.
 
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