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

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations

    3. 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 cant 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

    Attached Files:

  2. jcsd
  3. Nov 11, 2014 #2
  4. Nov 11, 2014 #3
    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 travelled 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.
  5. Nov 11, 2014 #4
    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).
  6. Nov 11, 2014 #5
    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).
    Last edited: Nov 11, 2014
  7. Nov 11, 2014 #6


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    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.
  8. Nov 11, 2014 #7

    Simon Bridge

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    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 lets you use the information you have, bearing in mind that some of the information may be red herrings.
  9. Nov 11, 2014 #8


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    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 [itex] mg \left( \frac{vh}{v-u} \right); v > u [/itex]]
    Last edited: Nov 11, 2014
  10. Nov 11, 2014 #9
    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........
  11. Nov 11, 2014 #10
    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.......
  12. Nov 11, 2014 #11
    pls can u explain that which non conservative force is in action here........??????
  13. Nov 11, 2014 #12

    Simon Bridge

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    ... what s the magnitude of displacement called?
  14. Nov 11, 2014 #13


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    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 Fnet = 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).
    Last edited: Nov 11, 2014
  15. Nov 12, 2014 #14
    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.

    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).
    Last edited: Nov 12, 2014
  16. Nov 12, 2014 #15
    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
  17. Nov 12, 2014 #16
    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.
  18. Nov 12, 2014 #17
    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
  19. Nov 12, 2014 #18
    No problem (:
  20. Nov 12, 2014 #19
    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......
    Last edited: Nov 12, 2014
  21. Nov 12, 2014 #20


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    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.
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