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Relationship of Power and Driving Force

  1. Nov 30, 2015 #1
    I've been teaching the Work, Energy and Power in UK A'Level Mechanics for some years without a problem. However, I got a question in class today which really made me think about my deeper understanding of the topic. I wonder if anyone can help with explanation of the problems below.

    The questions were in relation to problems involving a car being propelled by some kind of engine and experiencing a constant external resistive force.

    In this situation the equation Power = Driving Force x Velocity is used. If the power output of the engine is constant then this implies the Driving Force starts at a maximum value and gradually reduces as the velocity increases until the car reachs a maximum velocity and the car stops accelerating. At this pointy the driving force must be equal to the resistive force. When challenged I struggled to explain why the driving force is varying and further why is should start at a high value and then reduce (if anything it seems it should be the other way round). I tried to think in terms of the work done by the engine with some being converted to kinetic energy, as well as heat and sound, but this didn't help.

    This all got me thinking about the derivation of P=Dv I’ve always done this by differentiating work (written as force x distance) stating the force is assumed constant. However, in the example above it seems clear that D is some kind of function of t!

    I then started thinking about Power in terms of energy transfer and noted that for an idealised system where all work done by an engine is converted to Kinetic Energy, taking the rate of change of KE leads to the same formula (P=DV). So is the above due to some kind of implied modelling assumption?

    Can anybody help with a better explanation I can offer my inquisitive sixth formers.
     
  2. jcsd
  3. Nov 30, 2015 #2
    Do you have doubts about force being larger "in the beginning"? Or in lower gears?
    The force is larger when you start and decreases as you increase the speed.
    Why do you think it should be the other way around?
     
  4. Nov 30, 2015 #3
    Not thinking about anything as complex as a gear system. It seemed to me that the engine would need to use more engery a it warms up leaving less to provide ke to the system. For example is the lubrication not more effective as the engine heats? Also does the engining not have to overcome initial ineria when it starts, again leaving little energy to be transfered to ke?
     
  5. Nov 30, 2015 #4

    jbriggs444

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    It is precisely the gearing system which allows an engine running at maximum power (and therefore a fixed rotation rate) to operate at a range of velocities. It is precisely the gearing system which results in a reduction in driving force as velocity increases.

    The gearing system is not complex. It is simple. Far simpler than competing theories about lubrication efficiency as a function of temperature.
     
  6. Nov 30, 2015 #5

    A.T.

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    Without gearing, how do you keep the power constant at different speeds?

    This might help you:




    Nasu talks about force, you answer about energy. Try to keep the two apart
     
  7. Nov 30, 2015 #6
    OK, I think the penny has dropped about the necessity of a Gears in order for there to be a variable driving Force, but what about the second point taking the derivative of a fixed Force when it is in fact changing with time either because of a gear change, or perhaps with a continuously variable gearing?
     
  8. Nov 30, 2015 #7

    A.T.

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    If force is not constant, then the formula for work is an integral. The power formula doesn't change, except that F is a function of time.
     
  9. Nov 30, 2015 #8
    Thanks for that, this makes sense to me now. Great video, I use it my pupils tomorrow.
     
  10. Nov 30, 2015 #9

    russ_watters

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    Note that another contributor that has been glossed over is acceleration. Any force left over after accounting for drag and friction contributes to acceleration via f=ma.

    So if you had a constant power, the acceleration starts high and then drops as drag and speed increase.
     
  11. Dec 5, 2015 #10
    Here's an excel sheet to look at, its a trial and error algorithm for calculating top speed from predicted engine power and involving other factors such as drag coefficient rolling resistance etc., you can pre calculate the rolling resistance using the rolling resistance tab.
    remember the tan boxes are inputs and the v start figure should be greater than zero.
    have fun, I've used some example data but you can alter the inputs to suit yourself
    i wont be viewing this site again until Monday.
     

    Attached Files:

  12. Dec 5, 2015 #11
    Thanks for this, a couple of my pupils have interviews for Engineering coming up at Oxbridge, I'll share this with them - just the kind of thing they need as discussion points.
     
  13. Dec 10, 2015 #12
    by the way, the maths assumes that the drag co-efficient is the same in both directions.
     
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