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The effect of passenger walking on power consumption of escalator Watch

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    I have been thinking a lot on this question since i came across it in my revision notes. I got really confused as to whether the power of the escalator remain unchange, increase or decrease when it ferry walking passenger upwards.
    My assumption is that when you walk, you are exerting a downwards force onto the inclined escalator which in turn must generate more power in order to maintain its constant speed.
    To add on, if you walk on the escalator, you go up faster thus more work has to be done per second? but i am not sure about this as to whether this increase in work is supplied by the escalator or the passenger muscle or both.

    Any bright souls out there?
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    (Original post by haruchris)
    I have been thinking a lot on this question since i came across it in my revision notes. I got really confused as to whether the power of the escalator remain unchange, increase or decrease when it ferry walking passenger upwards.
    My assumption is that when you walk, you are exerting a downwards force onto the inclined escalator which in turn must generate more power in order to maintain its constant speed.
    To add on, if you walk on the escalator, you go up faster thus more work has to be done per second? but i am not sure about this as to whether this increase in work is supplied by the escalator or the passenger muscle or both.

    Any bright souls out there?
    That's a very good question

    Your first thought (about exerting downwards force and further on) seems correct to me.

    As for the work done, it's not that obvious. If the escalator can supply more power, it will do more work per second. Walking up, you also surely do some work. So the additional work per second is supplied both by yourself and the escalator (if its motor can do that).
    However, if you're asking about the work done to lift you up, then we have a problem Work is power times change in time, right? So the motor uses more power but the time of your movement is shorter. Let's consider it with the help of maths.
    If there is no force exerted by the man's walking, we have that the power supplied by the escalator is P=FV. If the man exerts force f on the escalator, it has to supply power P_1=\left( F+f\right) V to allow for constant speed. The work done by the escalator is power times the time to lift the man up: W_1=P_1t. The time needed for this is t=s/\left( V+v\right) where v is the man's own speed relative to the escalator. So the work done by the escalator is W_1=\left( F+f\right)\dfrac{V}{V+v}s. Normally the work done by the escalator would be W=Fs. So we can find when W_1>W:

    \left( F+f\right) \dfrac{V}{V+v}s>Fs

    FV+fV>FV +Fv

    \dfrac{f}{v}>\dfrac{F}{V}.
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    Let's first make sure we understand the difference between energy consumption and power (or average power) consumption.
    If the vertical height you travel from the bottom to the top is h, then the total energy required to get you there is mgh. Let's say it takes a time T to get you to the top if you stand still.
    If you stand still, all this energy is supplied by the escalator. If you walk, some is provided by you and some by the escalator. The total is always mgh.
    If, for example, you walk at the same speed as the escalator moves, you will get to the top in half the time and walk up half the total number of steps available. You do half the work and the escalator does the other half to get you to the top. ½mgh each.
    The energy the escalator needed was ½mgh, but you got there in half the time, so the average power developed by the escalator while you were on it was \frac{0.5 mgh}{0.5T}
    This is exactly the same as if you had stood still. Then it woud have been \frac{mgh}{T}

    This analysis ignores the energy required by the escalator to simple move itself even when there are no people on it. The values above would be the amount of energy required over and above this.
 
 
 
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