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Power required to drive up a hill

  1. Nov 1, 2014 #1
    • Forum guidelines require the use of the homework template. This thread has not been deleted due to already having relevant answers.
    Consider a 1200kg car cruising steadily on a road at 90km/h. Now the car starts to climb a hill that is sloped a 30 deg from horizontal . If the velocity of the car is to remain constant during climbing, determine the additional power required that must be delivered by the engine.

    KE=mv2/2

    velocity=9/3.6=2.5m/s

    KE= 1200*2.52/2=3750 J

    No sure where to go from here.
     
  2. jcsd
  3. Nov 1, 2014 #2

    BvU

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    Please use the template as was asked from you before. [edit] my mistake, but nevertheless.

    In that post you can find all you need, so there's no reason to be unsure.
    Look at the rate of change of the total energy and ask yourself where that has to come from.
     
  4. Nov 1, 2014 #3
    IMG_20141101_183728.jpg

    Assume the hill to be frictionless because we have no information about the coefficient of friction. So the force the car has to exert to go up the hill at constant velocity must be equal to 1200g( from newtons second law, because you dont want rhe car to acceleeate) .Using this value , you can calculate the work done by the car, and then the work done per second, because you know the velocity
     

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  5. Nov 1, 2014 #4
    PE=m*g*h


    h=cos30deg
     
  6. Nov 1, 2014 #5
    The difference is the (additional) power required to overcome the gravity force added by the incline ( m * g * sine ( 30 ° ) )
     
  7. Nov 1, 2014 #6

    Borek

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    That's some real steep road. 58% :))
     
  8. Nov 1, 2014 #7
  9. Nov 1, 2014 #8
    PE=1200*9.81*(sin 30)=5886 J
     
  10. Nov 1, 2014 #9

    BvU

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    Not really, Borek. Only 50% :)
     
  11. Nov 1, 2014 #10

    BvU

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    No. Wrong dimension.

    In fact, the earlier h=cos30deg had the same problem.
     
  12. Nov 1, 2014 #11
  13. Nov 1, 2014 #12

    BvU

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    1200 is kg, g is m/s2. Joules is kg m2/s2.
     
  14. Nov 1, 2014 #13
    YA stupid of me.
     
  15. Nov 1, 2014 #14

    BvU

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    This is going to be a long thread if we don't make some headway. Power is energy per time. Energy is force times length, so Power is force times length/time. All the ingredients are at hand now, so what's the answer (preferably in an expression, the numbers aren't all that interesting)
     
  16. Nov 1, 2014 #15
    Power= m*g*(sin30)*v

    P=1200*9.81*(0.50)*2.5=14715w, 147kw
     
  17. Nov 1, 2014 #16

    BvU

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    Almost. Another remnant (from the first post, this one! Don't write 9/3.6, instead: ) v = 90000 m / 3600 sec = 25 m/s. So the 14715 becomes 147150 and that is 147 kW
     
  18. Nov 1, 2014 #17

    Borek

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    There are two, slightly different definitions - one uses sinus, the other tangent. One yields 50%, the other 58%. My understanding is that the one based on tangent is more commonly used.

    Whichever we use - still steep :)
     
  19. Nov 1, 2014 #18

    Mark44

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    The only one I've ever seen used for the slope of a road is "rise/run", or tangent.
     
  20. Nov 1, 2014 #19

    BvU

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    Not so easy to determine which one is used for the slope.
    Sine has the advantage it's useful: odometer tells you how far you've climbed
    But I see a lot of "definitions" use the tangent, indeed. So slopes can be well over 100%, up to infinity !
     
  21. Nov 1, 2014 #20

    haruspex

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    The standard way to describe a hill on British road signs is "1 in N", where N refers to the hypotenuse, so N would be the cosec. I believe it is also used for railways.
     
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