Work Done by Man in Going Up Against the Motion of an Escalator

[sarthak sharma]

Homework Statement

an escalator is moving downwards with uniform speed u and a man of mass m is running upwards on it with uniform speed v. if height of escalator is h find the work done by man in going up the escalator

Homework Equations

The Attempt at a Solution

on my first shot at this question i thought that it should be mgh which is the change in his potential energy
but now i think that it should be wrong because he also has to do work against the motion of escalator.

i tried my best but could not solve it and also i can't understand the solution to it in my book...so i have attached a file for the same
i would be glad if someone can help me out at this

 

[sarthak sharma]

@collinsmark @nasu

 

[Jazz]

If u were equal to 0, then the runner would run a distance d (it's supposed to be the hypotenuse), and the work done would be just mgh, right?

But since the escalator has a velocity that is opposing to the motion of the runner, the upward net velocity is v - u (which is less than just v). And you can see that the 'effective distance' run will be greater as long as u > 0, u $\neq$ v and hence the work done will be greater too. The distance traveled is increased by a factor of v/(v-u) and if the distance is increase by that factor, so does the work done.

You can use the same approach used in the problem of going in a boat in a direction opposite to that of the current velocity: it will take you a longer time to go upwards than going downwards.

 

[nasu]

I am not sure what they mean here.
What kind of work?
A man can "do" a lot of internal work on his body parts, spend a lot of effort but without moving at all as a whole.
The work done against gravity is the same (for a given height difference), no matter how long is the actual distance traveled or how fast (assuming constant speed).

 

[Jazz]

I am not sure what they mean here.
What kind of work?
A man can "do" a lot of internal work on his body parts, spend a lot of effort but without moving at all as a whole.
The work done against gravity is the same (for a given height difference), no matter how long is the actual distance traveled or how fast (assuming constant speed).

That's is true, but since the man exerts a nonconservative force, the path should matter, shouldn't it?

If h were 0 (it's just a moving walkway) and he and I were in it: he moving a box against the walkway's motion and I just standing. To me, he will be moving the box with a force F through a distance d', which is greater than the distance d observed from the outside.

In the problem, the PE will be mgh independent of the path followed, but the work done by the man in the process should depend on the distance traveled.

In that regard, the first part of my first comment isn't correct since the force exerted by the person would exactly mgh only if he were moving straight up (climbing or something like that).

 

[collinsmark]

The work the man does on gravity is mgh. That much I think we can all agree on.

But that's not the end of the story. The question asks what is the work done by the man, which assumes his total work. The man also does work on the escalator.

Consider the extreme case where v approaches u. In that case the man is essentially powering the escalator; the escalator may not even need electricity in that case, it just needs the man -- the man's work on the escalator could even be used as the energy to bring other people up the opposite escalator one at a time. And that can keep going day and night since the man is always on the escalator.

 

[Simon Bridge]

Consider the extreme case where v approaches u. In that case the man is essentially powering the escalator; the escalator may not even need electricity in that case, it just needs the man -- the man's work on the escalator could even be used as the energy to bring other people up the opposite escalator one at a time. And that can keep going day and night since the man is always on the escalator.

You'd think being driven would make a difference to the mans work on the escalator wouldn't you - since there is work from the motor added to work from the man? ... However, an ideal escalator (frictionless and silent) does not require any work to keep it moving at a constant speed.

To walk in place - the man would move up a step, then the step moves down, with the man on it, so he walks another step up to where he started ... so accumulate lots of small mgh's of work, where h is the height of each step, over time ... except he could also be skilled enough to keep his center of mass in place and just work to lift his feet - and we don't have information for the mass of the feet!

To do this problem requires figuring out a model that let's you use the information you have, bearing in mind that some of the information may be red herrings.

 

[collinsmark]

You'd think being driven would make a difference to the mans work on the escalator wouldn't you - since there is work from the motor added to work from the man? ... However, an ideal escalator (frictionless and silent) does not require any work to keep it moving at a constant speed.

That's only true assuming there is yet another external force supporting the man's weight. (Imagine the man is supporting himself on some stationary rails, although his legs are allowed to move.) In that case, there is no work required.

But if the man is supporting his own weight [with his legs, on the escalator, in the absence of any external, supporting apparatus], he is doing work on the escalator (i.e., the escalator's "motor" is acting as a generator), keeping it moving at a [non-zero,] constant velocity. Otherwise, the man, and the escalator, would accelerate.

[Edit: edited for minor clarifications. And btw, I'm positing that the answer worked out in the attachment (the book's solution) is correct. That is, the work done by the man is mg \left( \frac{vh}{v-u} \right); v > u]

 

[sarthak sharma]

[Edit: edited for minor clarifications. And btw, I'm positing that the answer worked out in the attachment (the book's solution) is correct. That is, the work done by the man is mg(vhv−u);v>u mg \left( \frac{vh}{v-u} \right); v > u ]

pls can u explain how u got the answer as stated above because i m unable to get that using all i could and i also don't get the method mentioned in book...

i also agree that the work done by man should be greater than mgh as he needs to do more work against the motion of the escalator but i m unable to relate this relative velocity to work done by man...

 

[sarthak sharma]

If u were equal to 0, then the runner would run a distance d (it's supposed to be the hypotenuse), and the work done would be just mgh, right?

But since the escalator has a velocity that is opposing to the motion of the runner, the upward net velocity is v - u (which is less than just v). And you can see that the 'effective distance' run will be greater as long as u > 0, u $\neq$ v and hence the work done will be greater too. The distance traveled is increased by a factor of v/(v-u) and if the distance is increase by that factor, so does the work done.

You can use the same approach used in the problem of going in a boat in a direction opposite to that of the current velocity: it will take you a longer time to go upwards than going downwards.

i just wanted to know that how work done is related to distance because as far as my knowledge goes we relate work done with displacement...
if there is anything different for this case can pls explain it a bit more...

 

[sarthak sharma]

That's is true, but since the man exerts a nonconservative force, the path should matter, shouldn't it?

pls can u explain that which non conservative force is in action here...?

 

[Simon Bridge]

i just wanted to know that how work done is related to distance because as far as my knowledge goes we relate work done with displacement...

... what s the magnitude of displacement called?

 

[collinsmark]

pls can u explain how u got the answer as stated above because i m unable to get that using all i could and i also don't get the method mentioned in book...

Consider that each step on the escalator amounts to 1 unit* of displacement in the "up" direction (we can ignore the horizontal direction for this problem since there are no forces in that direction). So then it all amounts to "how many steps does the man take in getting from 0 to h (from the point of view of somebody watching from the ground)?

*(you may think of it as each step being 1 meter tall; if so, pretend that the man is unusually tall and flexible.)

Since nothing is accelerating, you know that F_net = ma = 0, which implies that the normal force from the escalator is equal and opposite the man's weight. So the total work done by the man is W = Fd, where d is the number of steps the man takes (and where each step is one unit of vertical displacement).

 

[Jazz]

i just wanted to know that how work done is related to distance because as far as my knowledge goes we relate work done with displacement...
if there is anything different for this case can pls explain it a bit more...

Distance is referred to the magnitude of displacement. But in this particular case it's necessary to talk about distance traveled. It was argued that the force exerted by the man is nonconservative, and hence the path (not just the $\Delta d$ between two points) must be taken into account. Consider that:

- In the best scenario, the work done by this force, $W_{nc}$, will be equal to the $PE_g$. Otherwise it will be greater.
- The distance traveled (let call it $d$) is $d =\bar{v}t$, where $\bar{v}$ is the average velocity. Then $t = \frac{d}{\bar{v}}$

The net, average velocity was $v-u$. It follows that:

$W_{nc} \geq PE_g$

$F_{nc}d \geq mgh$

$F_{nc}\frac{d}{v-u} \geq mg\frac{h}{v-u}$ <-- We want to find for how long the force was exerted (without writing $t$ explicitly). Look at the right-hand side. You end up with units of momentum. $\frac{d}{v-u}$ is the time he was running in the escalator (another way of viewing this is by looking at the step the man took, as collinsmark says).

$F_{nc}\frac{d}{v-u}v \geq mg\frac{h}{v-u}v$ <-- Now we want to find the distance traveled by the man. We multiply the time he spent running (written in terms of $d$ and $v-u$) by the (average) velocity of the man $v$.

The equation is just telling us that there is no way that the work done is equal to $mgh$. You need to either climb straight up a height $\frac{h}{v-u}v$ at constant velocity, or to consider that the work done also went to other forms of energy.

pls can u explain that which non conservative force is in action here...?

The force the man exerts going up the escalator. Let suppose it's a conservative force. Then, Conservation of Mechanical Energy should apply:

$KE_i + PE_i = KE_f + PE_f$

$KE_i = KE_f + PE_f$ <- take $h_i = 0$

$KE_i = PE_f$ <-- Suppose at the top his $v = 0$.

Now pick an arbitrary mass (of a person), a fixed $h$ and vary $v_0$. The equation doesn't hold. By Conservation of Energy that can't be right. The difference energy input/output must go somewhere. Then we conclude that the force exerted should be nonconservative:

$W_{nc}= PE_f + OE_f$

$F_{nc}d = PE_f + OE_f$ <- some of the energy going up goes into other form of energy (OE).

If $F_{nc}d = PE_f$, means that $OE_f=0$, which in turn means that the man doesn't feel tired, hot, doesn't need to take a breath, etc (is a superhero).

 

[sarthak sharma]

Distance is referred to the magnitude of displacement.

don't you think that this applies in only some cases but not in all...like if we go 30m towards north then back 20m towards south so here distance is 50m but displacement is 10m towards north

 

[Jazz]

Displacement is a vector (has magnitude and direction).
Distance is just the magnitude of the displacement.
Distance traveled is the path taken between the starting and ending points.

Displacement would be 10m towards north.
Distance would be just 10m.
Distance traveled would be 50m.

 

[sarthak sharma]

Displacement is a vector (has magnitude and direction).
Distance is just the magnitude of the displacement.
Distance traveled is the path taken between the starting and ending points.

Displacement would be 10m towards north.
Distance would be just 10m.
Distance traveled would be 50m.

okay in that way i accept it...it was my mistake as i was considering that by referring to "distance" you mean distance travelled...:P

 

[Jazz]

No problem (:

 

[sarthak sharma]

i took this problem to one of my seniors and he gave me a different way to solve it which seems to be a much easier method...
it is as stated below...i would like to know about what you all think about this solution...

relative velocity=v-u ((this we all know and agree upon))
if u=0 i.e. escalator is at rest then time taken to reach at top is t = h/v and displacement would be d=vt=h
and when u≠0 then time taken to reach at top is t' = h/(v-u)
so extra time taken to reach at top is t'-t = uh / v(v-u)

thus extra displacement in this extra time if escalator was stationary is d'=v(t'-t)= uh / (v-u)
so total displacement D = d+d' = h + uh / (v-u) = vh / (v-u)

and hence work done is mgvh / (v-u)

please look at this solution and notify me if anyone of you finds any error in it...

 

[PeroK]

i took this problem to one of my seniors and he gave me a different way to solve it which seems to be a much easier method...
it is as stated below...i would like to know about what you all think about this solution...

relative velocity=v-u ((this we all know and agree upon))
if u=0 i.e. escalator is at rest then time taken to reach at top is t = h/v and displacement would be d=vt=h
and when u≠0 then time taken to reach at top is t' = h/(v-u)
so extra time taken to reach at top is t'-t = uh / v(v-u)

thus extra displacement in this extra time if escalator was stationary is d'=v(t'-t)= uh / (v-u)
so total displacement D = d+d' = h + uh / (v-u) = vh / (v-u)

and hence work done is mgvh / (v-u)

please look at this solution at notify me if anyone of you finds any error in it...

I think the key to this problem is the proportionality of the work being done against height gained. The original solution you posted simply states that it is proportional: v/(v-u). The solution above introduces time but then assumes that the escalator is vertical (can you see why?).

It all boils down to: you are doing work at a rate v, but only progressing at v-u, so the total work is v/(v-u) times what you actually gained.

If you want to do this "properly" and prove that the proportionality actually holds (this might be a good idea). Then, you need to introduce the angle of the escalator. Then, calculate the rate of work and height gain and show that the angle cancels out in the end.

 

[sarthak sharma]

please look at this solution and notify me if anyone of you finds any error in it...

@collinsmark hey can you please have a look at the solution as i mentioned above and add your comments on it...

 

[Jazz]

i took this problem to one of my seniors and he gave me a different way to solve it which seems to be a much easier method...
it is as stated below...i would like to know about what you all think about this solution...

relative velocity=v-u ((this we all know and agree upon))
if u=0 i.e. escalator is at rest then time taken to reach at top is t = h/v and displacement would be d=vt=h
and when u≠0 then time taken to reach at top is t' = h/(v-u)
so extra time taken to reach at top is t'-t = uh / v(v-u)

thus extra displacement in this extra time if escalator was stationary is d'=v(t'-t)= uh / (v-u)
so total displacement D = d+d' = h + uh / (v-u) = vh / (v-u)

and hence work done is mgvh / (v-u)

please look at this solution and notify me if anyone of you finds any error in it...

You can take a look at what @PeroK has written. You got the correct answer, but making wrong assumptions. And if you are more careful, the correct answer isn't quite correct either.

The length of the slope is $d$ and $d = h$. Then the work done is $mgd$, but this is equal to $mgh$.

The same assumptions are being made by the problem if you set $u = 0$ in the given answer. It's assuming that whatever the amount of work done going upward was, when the velocity was $v$, it was fully stored as $PE_g$.

Can you see what underlying assumptions are being made by setting $W = mgh$?

 

[collinsmark]

i took this problem to one of my seniors and he gave me a different way to solve it which seems to be a much easier method...
it is as stated below...i would like to know about what you all think about this solution...

relative velocity=v-u ((this we all know and agree upon))
if u=0 i.e. escalator is at rest then time taken to reach at top is t = h/v and displacement would be d=vt=h
and when u≠0 then time taken to reach at top is t' = h/(v-u)
so extra time taken to reach at top is t'-t = uh / v(v-u)

thus extra displacement in this extra time if escalator was stationary is d'=v(t'-t)= uh / (v-u)
so total displacement D = d+d' = h + uh / (v-u) = vh / (v-u)

and hence work done is mgvh / (v-u)

please look at this solution and notify me if anyone of you finds any error in it...

Yes, that leads to the correct answer, but it's kind of a round-about way of doing it. But it get's the job done.

You might consider this method instead: Rather than calculating the "extra" time taken, simply find the total time is takes him to reach the top of the escalator.

The next part is to calculate his total vertical distance traveled, measured from the Galilean frame of reference of the escalator stairs, where is is traveling at speed v.

Others have mentioned the importance of taking the angle of the escalator's slope into consideration. While there's nothing wrong with doing that, I think it's possible to reduce v and u, and their proportionality, directly to vertical velocities with nothing more than some handwaving (such as what your book's solution did).

 

[collinsmark]

By the way, in an attempt to clear up some confusion, realize that who does work on what is always dependent on the Galilean frame of reference chosen. Generally speaking you can get different intermediate answers for different frames of reference.

In my previous post, I suggested using the Galilean frame of reference of the escalator stairs, on which the man is traveling at speed v, because the answer is asking for the total work done by the man. And that's an easy way to get that answer.

But if you wanted to, you could instead choose the Galilean frame of reference of the building (i.e., ground, floor, etc.). In that case, the man does work on gravity (which we established is mgh), but since the man also exerts a force on the escalator stairs, over a vertical difference traveled by the escalator stairs (and in this case you measure the vertical distance the stairs travel with respect to the building [ground, floor, etc.]), there is work done on the escalator too. So you need to combine those for the final answer.

You'll get the same answer either way in the end. The intermediate steps are just a matter of how you look at it.

 

[collinsmark]

In case that wasn't clear, consider a third frame of reference to use in order to solve this problem: Choose the Galilean frame of reference of the man himself. This is the most direct way to solve the problem.

In this case, the frame of reference is chosen such that the man is standing still.

The stairs move past the man at a speed of v. The building (and ground and Earth) move relative to the stairs with a speed u.

The building itself (along with the ground and the Earth) is moving in this frame of reference. By the time the top of the escalator reaches the man, the building (and ground and Earth) will have moved down a vertical distance h.

In this situation, the man is applying a force on the stairs and they move a certain distance relative to himself. Thus you can simply find W = \vec F \cdot \vec d once you calculate that vertical displacement of the stairs (relative to the man).

Once again, you'll end up with the same final answer as with the other methods.

 

[haruspex]

relative velocity=v-u ((this we all know and agree upon))

No, that's backwards. Although it is not clearly stated, the given v is the speed, measured upwards, relative to the escalator (to get that answer). Upwards 'absolute' speed (i.e. relative to the ground) is v-u.

 

[Neeraj Chandran]

There is no need of making it complex. Actually it is very simple. Imagine an escalator with a height 'h' and it is constant even if the belt is moving in the opposite direction. Let the velocity of belt be 'u' and velocity of man be 'v'.
The velocity with which the man climbs the escalator is ' v-u ' (because the lift is moving in the opposite direction. So the velocity will be lower compared to v.
We know that time=distance/speed, therefore the time taken to get to the height h is h/(v-u).-------------->{1}
We know that he covers the distance h but at that time, as the escalator is continuously moving backward, he covers the distance above the escalator belt with velocity v .i.e.
Distance=velocty×time
d=v×(h/v-u) from {1}----------->{2}
As he is moving with constant velocity net force is 0. That is a force 'mg' acts on the upward direction. Therefore
Work=force×displacement
W=(mgvh)/v-u from {2}

 

[gneill]

@Neeraj Chandran : you're responding to a homework thread that is over four years old. Your contribution is unlikely to be relevant to the Original Poster at this time. Just thought you should be aware.