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Work on a car that is skidding to a stop

  1. Mar 13, 2015 #1
  2. jcsd
  3. Mar 13, 2015 #2

    Merlin3189

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    No, because it hasn't. I think you misread it.

    "The work done by friction on the car is related to the initial kinetic energy of the car. The work-energy relationship is often related by the equation

    KEi + PEi + Wext = KEf + PEf "

    Which is different from what you said.

    Then they point out that the PE is the same throughout, assuming the car is going along level ground.

    "Since the potential energy of the car is the same in the initial state (before braking) as the final state (after stopping),"

    And then they cancel all the irrelevant and null terms

    " each term can be canceled from the above equation. And since the car is finally stopped, the KEf term in the equation is zero."
    So that KEi + Wext = 0 though they don't bother to explain, just substitute straight in

    "Thus, the equation becomes 0.5*m*v2 + F*d*cos(180) = 0."

    Basically KE = Work done by skidding


    In the absence of external gain or loss of energy, such as a ball rolling (or sliding) along a frictionless surface in the absence of air resistance, then KE + PE = KE + PE , so if it stops, it must have gained PE.
    But here the lost KE went into the friction.
     
  4. Mar 13, 2015 #3
    Thank you for the explanation. So would there be work done either negative or positive, or is work = 0?
     
  5. Mar 13, 2015 #4

    Merlin3189

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    Work is done by the car - "burning" rubber! ie. breaking up the tyres, breaking up the road surface, heat from friction on the road and in the brakes and tyres, wear on the brakes, noise,

    Imagine you were pushing the car along with its brakes on! You'd have to do a heck of a lot of work. Well the car is doing that work as it skids along: which is why it stops pretty quick when all its KE is used up in doing that work.
     
  6. Mar 14, 2015 #5
    Then I do not understand why the answer to question 6. a) is 'none' on this worksheet: http://www.hartlandhighschool.us/subsites/Andrew-Kartsounes/documents/AP%20Physics%20C/5%20Energy/Masters/Energy%20Packet%202%20Key.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  7. Mar 14, 2015 #6

    jbriggs444

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    The question in the reference asks for the "work done on the system". The implication is that this means work done by forces exerted from outside the system. An interaction between two objects within the system (i.e. the car and the road) do not count.

    That said, I am not happy with the response given in post #4 above. Nor am I completely happy with the answer in the reference.

    The car is not doing work on the road. The car is not doing any work at all. It is applying a force on a motionless object (the road). No matter how much force it exerts, the work it does will be zero. The road is not doing positive work on the car. The dot product of the force exerted by the road and the motion of the car is negative. The road is doing negative work -- removing energy from the car.

    The net effect of this interaction is to remove kinetic energy from the system. Even though it is an internal interaction, it reduces the bulk kinetic energy of the system. Of course, since it is an internal interaction, it does conserve energy. The lost kinetic energy shows in other forms -- primarily as heat. Still, this internal interaction has done net negative work on the system. It would be reasonable to say that negative work has been done "on the system" even though no external agency is involved.
     
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  8. Mar 14, 2015 #7

    Merlin3189

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    I must admit, I don't understand any of question 6. All the systems seem to be closed and any work done is exchanged between parts of the system.
    I am saying that in 6a the car is doing work on the pavement and the earth. It is applying a force to them and is moving in the direction of the force. I'm not familiar with the idea of negative work, but I assume the earth and pavement are doing the negative work, since the movement is in the opposite sense to the force they are applying.
    Work is the dot product of the force vector and the movement vector, so
    work = |force| x |distance| x cos(angle between force vector and movement vector)
    Since cos(0) = 1 you get a positive value for work when the directions are the same
    and cos(180 deg) or cos(π rad) = -1 you get a negative value for work when directions are opposite
    (and if they are perpendicular, cos(90 deg) or cos(π/2) = 0 , so you get zero work .)
     
  9. Mar 14, 2015 #8

    jbriggs444

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    That is not correct. The work done by a force is computed as the product of the force times the distance moved by the object on which the force acts. The road is not moving, so no work is done. [There is another layer of complexity if one deals with rotating or non-rigid objects, but the road is rigid and non-rotating, so we need not deal with that].

    Irrelevant. It is the motion of the target of the force that matters, not the motion of its originator.
    Yes. Correctly reasoned.
     
  10. Mar 14, 2015 #9

    Merlin3189

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    I can't really dispute this as I'm not an authority on this subject, but that is not my understanding.
    If I push on an object, such as the road, and my hand moves in the direction I'm pushing, I do not know whether the road is moving or not, I only know I'm doing work. If the road moves (maybe it is a rolling road in a garage) the work is (largely) transferred to KE in the road. If the road doesn't move, the work is (largely) transferred to heat in the road. In both cases, if the force is the same and the movement of the point of application is the same, I do the same work.
    (WikiP says, "In physics, a force is said to do work if, when acting on a body, there is a displacement of the point of application in the direction of the force." While I must admit WikiP is hardly authoritative, that does agree with what I learnt about this.)

    If you hold to the notion that energy (such as the heat of friction) can appear where no work is being done, are you not breaking the conservation of energy?
     
  11. Mar 14, 2015 #10

    A.T.

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    Your hand will heat up as well, eventually more than the road. Your force is not doing work on the road in this case.

    No, because the road is doing negative work on your hand. That's where the heat energy comes from.
     
  12. Mar 14, 2015 #11

    Merlin3189

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    You may have a point there. I am doing work, some on the road and some on my hand?

    Can the road be doing negative work on me, yet I not be doing positive work on the road? In what form does the negative work come from me? Does it come as work or as heat?
     
  13. Mar 14, 2015 #12

    Merlin3189

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    The difference between my point of view and that of authoritative sources seems to be, whether the object I'm pushing has to move or not.
    My position is simply that work is the dot product of the force I apply and the distance the point of application moves, irrespective of how or why that point of application moves.
    The generally stated position seems to be work is the force applied to an object times the distance the object moves in the same direction as the force.

    My general contention is that, since a physical object cannot know whether the force it is applying, is pushing is moving object or not, nor indeed whether it is even pushing an object (eg. electromagnetic forces), then it cannot matter whether an object moves.

    However, in the specific case pushing against friction, even though the nominal pushed object does not move, the atoms/molecules of which it is made do move and then spring back, converting tiny amounts of linear KE into random vibrational KE (apart from those pieces, which I mentioned in my first post, which are torn asunder doing work against their inter atomic/molecular bonds.)
     
  14. Mar 14, 2015 #13

    A.T.

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    No work is done on a static object. To model heating up using the work principle you would have to model the individual road atoms which are moving when your hand atoms apply a force to them.

    No, the hand cannot do work on the hand.

    Yes, and the difference in energy is dissipated at the interface.

    Work is F dot s, not heat.
     
    Last edited: Mar 14, 2015
  15. Mar 14, 2015 #14

    A.T.

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    You have to consider a physical point of application (a bunch of atoms) not just a geometrical. point.

    Yes, if you model all that detail you will have work being done on road atoms.
     
  16. Mar 14, 2015 #15
    I am currently in Honors Physics with a pretty poor teacher so I am quite lost after reading all of those different point of views.

    I read from above that there can be negative work while there can also be negative work? I may need to watch some videos so that my questions can come from a more educated standpoint.
     
  17. Mar 14, 2015 #16

    Merlin3189

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    My apologies Physi1csStud3nt. I'm afraid I got sidetracked into an abstruse argument with the other respondents.
    Please ignore my posts here and concentrate on theirs, which appear to agree with the majority of textbooks.

    Do however watch videos and look at some of the websites like physicsclassroom, hyperphysics, bbc bitesize, etc which do have some helpful material which should be quite intelligible to you.
     
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