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Homework Help: Person on moving stairs (Potential and kinetic energy)

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    The stairs (moving stairs like the ones in a mall) are moving with the vertical component of vs=1,5 m/s. What power should a man with a mass of m=80kg use to climb up to a height h=12m in t=5s using the stairs?


    2. Relevant equations
    g=9.81

    Book solution:
    [tex] P=\frac {mgh_p}{t} [/tex]

    [tex]h_p=h-v_st [/tex]

    [tex]P=\frac {mg(h-v_s t)}{t} [/tex]


    They calculated the height which the stairs take the man to, from which they got the height the man has to travel on his own. They used that height to calculate work.

    3. The attempt at a solution
    The average speed the man should go (with the speed of the stairs) is:
    [tex] v_{tot}=\frac {h}{t} [/tex]
    from which i can get the speed which the man should be going on his own.

    [tex]v_{man}=v_{tot}-v_{stairs} [/tex]

    So the kinetic energy of the man traveling at speed vman is:

    [tex]E_{k-man} = \frac {1}{2} m v_{man} [/tex]

    That is also the work the man does since his kinetic energy was 0 at the bottom of the stairs.

    So power should be:

    [tex]P=\frac{E_{k-man}}{t} [/tex]

    I don't get the same result as they do.

    My other assumption was that:

    [tex] E_p=E_{k-stairs}+E_{k-man} [/tex]
    [tex] mgh=\frac{1}{2}m(v_{man}^2+v_{stairs}^2) [/tex]

    From which I get that:
    [tex] v_{man}=\sqrt {2gh-v_{stairs}^2 [/tex]
    form which it turns out that the man's speed should be grater than the h/t.

    Where am i wrong?
     
    Last edited: Mar 7, 2010
  2. jcsd
  3. Mar 7, 2010 #2
    The kinetic energy that the man gets isn't important. He starts out at rest, and ends at rest as well. He has to do a little bit more work at the start to accelerate, but can do less at the end, as he can convert his speed into potential energy, but the average power is equal to what your book says.
     
  4. Mar 7, 2010 #3
    I didn't really understand you. Could you point out to which of my equations is incorrect?
     
  5. Mar 7, 2010 #4
    Everything from the point where you first mentioned kinetic energy
     
  6. Mar 8, 2010 #5
    I'm not seeing why isn't the kinetic energy of the man:

    [tex]
    E_{k-man} = \frac {1}{2} m v_{man}
    [/tex]

    for the work he does by himself

    [tex]
    E_{k-stairs} = \frac {1}{2} m v_{stairs}
    [/tex]

    for the work the stairs are doing for him.

    Could you explain that again please?
     
  7. Mar 8, 2010 #6
    The kinetic energy isn't relevant, because the man can have a speed of 0 at both the
    start and the end of the stairs. Only the potential energy that he gains is important.
     
  8. Mar 8, 2010 #7
    The kinetic energy isn't relevant, because the man can have a speed of 0 at both the
    start and the end of the stairs. Only the potential energy that he gains is important.
     
  9. Mar 8, 2010 #8
    Lets just observe the situation of a man on static stairs.

    At start he has some energy chemical energy stored in him. Then he starts to move, so some of his chemical energy is transformed into kinetic. And at the and, as he reaches the top, his kinetic energy transformed into potential.

    What's wrong with this?
     
  10. Mar 8, 2010 #9
    Nothing, but since there is no net gain in kinetic energy, there is no need to consider it.
     
  11. Mar 8, 2010 #10
    So why doesn't :

    [tex]
    P=\frac{E_{k-man}}{t}
    [/tex]

    than give the correct solution?

    [tex] P=\frac { \Delta E}{t} [/tex]

    Where [tex]\Delta E[/tex] is the change of the man's kinetic energy at starting. It is 0 while he is standing and rises to Ek when he starts moving.

    I understand the way it's solved in the book, i just want to get the same answer the other way around or see why isn't it possible to do so. :)
     
    Last edited: Mar 8, 2010
  12. Mar 8, 2010 #11
    Thank you willem2 for your previous responds.

    Is there someone else who could explain this in other words?
     
  13. Mar 8, 2010 #12
    mgh=\\frac{1}{2}m(v_{man}^2+v_{stairs}^2)

    For a start this assumption is wrong (just by putting in the numbers) and also as there is no set speed for vman, he could start off fast and finish slowly. Work can be defined as change in kinetic energy, but there is not necessarily a net gain of kinetic energy between the top and the bottom. maybe it would help to think that the man does work against gravity, and the force of gravity does work against the man but this is negative (and becomes the potential energy)

    The change in potential energy is the work done by the man against the force of gravity, and W=Fd. The average speed can be taken into account if you differentiate that to get P=Fv = mgv=mg(h-vst) as stated by the book, but you cannot use the average speed to work out a value for kinetic energy which you can then equate with work done (change in potential energy).

    Sorry if this isn't very clear, but do some research into it and see if there's a better explanation elsewhere online!
     
  14. Mar 11, 2010 #13
    Thanks for the effort.

    I think key for understanding why my assumption is incorrect is to understand why i can't use the average speed for the calculation of kinetic energy.

    If i understood you correctly, i could use [tex]

    E_{k-man} = \frac {1}{2} m v_{man}

    [/tex] but only in the case that i knew v_man at the beginning of the stairs?
     
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