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Work done by friction

  1. Apr 13, 2015 #1
    If I stand up on an accelerating bus, static friction between my shoes and the bus will do work.
    - I suppose it is negative work on the bus, and when velocity is constant it does no work at all, am I right?
    - does it do any work on me?
    - if I do not stand on my feet , but sit leaning my back onto a chair, the bus is pushing me. It is doing positive work on me, what about friction? is it zero?
    -if it is zero, why? the result is different, me moving at same speed of the bus, why is work done different?

    Thanks.
     
  2. jcsd
  3. Apr 13, 2015 #2

    Delta²

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    a) yes you are right. Only that when the velocity is constant, the work is zero because the static friction is also zero
    b) yes when bus is accelerating the static friction between shoes and bus floor does positive work on you that increases your kinetic energy.
    c) yes when bus is accelerating the force from the seat is pushing you and is doing positive work on you. The work from friction will be little or zero, as the main force that accelerates you is the force from the seat, the friction force is small (or even zero) compared to the force from the seat.
    d) it is zero because when you seat you arrange your body in such a way that the interaction between you and the bus is mainly done via the force from the seat, and not from the friction between your shoes and the bus floor. The friction at your shoes will be very little or none.
     
  4. Apr 13, 2015 #3

    A.T.

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    Depends on the reference frame. In the rest frame of the bus it does no work.

    Yes, in the ground frame for example.

    Yes.

    The friction will be more at your butt, than at the shoes. And there will be normal contact forces from the back of the seat.

    If you accelerate in the same way in both cases, then the total work on you by the bus will be the same.
     
  5. Apr 13, 2015 #4

    sophiecentaur

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    I find it hard to understand why people are worried by the whodunnit thing with friction and work. The friction is not a 'source' of energy, it is merely a set of contact forces between two objects (you can include fluids in this). Friction may be deemed 'responsible' for slowing things down, speeding them up or keeping them from moving at all. But it's only a shorthand name and you can put an arrow on a diagram with a label 'friction', just as you can put 'weight' or 'thrust'.
     
  6. Apr 13, 2015 #5
    Of course all the problems on the subject refer to the road frame.

    If I got it right, you are saying that when I am standing,
    - friction is doing negative work on the bus and positive work on me?
    - And the amount, I suppose is identical. If so , should we count it twice?
    suppose my weight is 50Kg , mu is 0.5 and d=50m, then F =mu mg =250N and the work done on the bus and on me is 2500 or 5000 J ?
    - And when I am sitting the bus is doing work on me, it can't be still 2500, right?
    - if we neglect the butt, (suppose I am tied to the seat) , what is mu between my back on the seat, 0, 1 or greater?

    And what do you make of the last answer, of course the source of energy is only the engine of the bus, isn't it?

    please address all questions so I must not repeat myself

    Thanks
     
  7. Apr 14, 2015 #6

    Delta²

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    yes friction is doing negative work on bus and positive on you. The total work of friction (betweem your shoes and the bus floor) for the system (bus+you) is zero. As you can see we dont count it twice, we take -W+W=0. Just notice that when we say friction here we actually mean an action-reaction pair . Action is the friction force from the bus floor to your, reaction is the friction force from your shoes to the bus floor.

    When you sitting tied to the seat ( so we neglect anything that happens between your butt and the seat, we only care what happens between your back and the seat) the bus is doing work on you but its not the work of friction between your back and seat(because this friction would be perpendicular to the displacement of the bus, the mu between your back and the seat can by anything and the friction between your back and the seat can be anything but this friction does not do work), it is the work of the normal force between your back and the seat . The normal force from the seat to your back is normal to the surface of your back but is not normal to the velocity (or displacement) of the bus, it is on the same direction with the velocity of the bus.

    yes, the source of energy is the engine of the bus. The energy provided by the engine of the bus is transformed to kinetic energy of you and the bus, via the work of internal forces (and the friction between your shoes and the bus floor, or the normal force between your back and the seat is an internal force).
     
    Last edited: Apr 14, 2015
  8. Apr 14, 2015 #7
    Thanks, Delta, can we put some numbers in it to make it clear, and could you write formulas in nice format?

    So, the engine is doing all the work P = M (B+me) * a (5) * d (50 m) = and friction is doing -W on the bus and + W on me W= mu m g = 0.5*50*10 N*50m= 1250 J?

    Now what I still do not understand is this:
    - if I stand, you say +W on me = -W on the bus so the engine is doing 1250J of work because nu = 1/2,
    - if I sit, the engine is doing same work on the bus but on me W = m 50* a 2* 50 m = 2500 J , double work?
    what is wrong?

    How do we get the work P done by the engine?
     
  9. Apr 14, 2015 #8

    Delta²

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    No the engine doesnt do only 1250, it does 1250J +the kinetic energy of the bus. However there is a problem in this consideration because we consider the friction force constant, while in fact the friction force changes in such a way as to keep you in the same velocity with the bus floor.

    A more proper way to put data in the problem will be that from an acceleration from 0 to say 10m/s, the work done from the friction (shoes-bus floor) or from the normal force (seat-back) to you will be W=DK=(1/2)m10^2, while the kinetic energy of the bus will be (1/2)M10^2, and the work of engine will be (1/2)(M+m)10^2. The net work on the bus will be (work of engine - work of friction)=(1/2)M10^2. That is ofcource neglecting air resistance and the loss of energy due to friction of bus tyres with the ground.
     
    Last edited: Apr 14, 2015
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