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Homework Help: Work Done yadda yadda These problems are killing me

  1. Oct 26, 2003 #1
    Ok, I'm having trouble with yet another work-related problem. Here it is:

    A traveler at an airport takes an escalator up one floor. The moving staircase would itself carry him upward with vertical velocity component 0.8 m/s between entry and exit points separated by height 10.0 meters. However, while the escalator is moving, the hurried traveler climbs the steps of the escalator at a rate of 2.8 steps per second. Assume that the height of each step is 0.3 meters.

    a) Determine the amount of work done by the traveler during his escalator ride, given that his mass is 80.0 kg

    b) Determine the work the escalator motor does on this person.

    For a, I summed the man's and the escalator's velocities and used the W = Fd equation. I'm not sure how to go about b.

    I was wrong for a, so I'm really up a creek on this one. Any help is greatly appreciated.
  2. jcsd
  3. Oct 26, 2003 #2


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    You did? Even though W= Fd has nothing to do with the velocities?
    It doesn't matter how FAST you apply a force, only that you apply it.
    I would have been inclined to note that the total work done is the
    change in potential energy. You are given that the man's mass is 80 kg. and that he has moved up a vertical distance of 10 m: change in potential energy= work done= (80)(9.81)(10)= 7848 Joules.

    For B, look at the work the man has done. HERE'S where the speeds come in. The elevator is moving up with a vertical speed of 0.8 m/s and the man is walking up at a vertical speed of 2.8 STEPS per second while each step is 0.3 m so that is 0.84 m/s, for a total vertical velocity of 1.64 m/s. The point of that is that it will have taken him 10/1.64= 6.10 seconds to go up the elevator.

    Now you can finish the problem in either of two ways:

    Directly: Since the elevator is moving upward at 0.8 m/s, it will have carried him up 0.8(6.10)= 4.88 m (he walked up the other 5.12 m)
    and thus did work of (80)(9.81)(4.88)= 3830 Joules on the man.

    Indirectly: The man was walking upwards at 0.84 m/s so in the 6.1 seconjds, he walked up (.84)(6.1)= 5.12 m (we saw that before) and thus did (80)(9.81)(5.12)= 4018 J of work himself. The elevator did the other 7848- 4018= 3030 J of work.
  4. Oct 26, 2003 #3
    Thank you a thousand times for your help. Really I don't know what I was thinking, using that equation. I should learn to consider if a solution makes sense before I apply it.

    Thanks again!
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