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Minimum Power of the engine on a car climbing a hill

  1. Aug 27, 2011 #1
    Hi, Can someone please solve this question for me, including step by step instructions ?
    I can solve the ones where the mass of a car is given, but can't do this one.

    1. The problem statement, all variables and given/known data

    Setting off from rest the car accelerates uniformly and climbs a steep hill which has a gradient of 1 in 3. It accelerates to 22m/s while travelling a distance of 100m up the incline.
    What is minimum power of the engine?

    2. Relevant equations
    None given in the question.
    P= F x v

    3. The attempt at a solution
    I keep on getting wrong answers ;(
     
  2. jcsd
  3. Aug 27, 2011 #2

    PeterO

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    Usually when a quantity is not given, it is because it is not necessary. To see if that is the case here [or perhaps an omission] work the problem assuming the car has mass, say, 900kg and 1400kg. [you said you cna do these problems when you know the mass]

    If you get the same answer each time, it means the mass really was unnecessary. If you get a different answer, then the mass was important.

    Perhaps the answer if of the form XX kW/Tonne?

    And re-read the question carefully - perhaps the mass of the car was actually given and you didn't notice.
     
  4. Aug 27, 2011 #3
    This is part C of the Exercise, mass was given fort part B and it was 600kg.


    The answer is 36.2kW

    However I get 42kW :(
     
    Last edited: Aug 27, 2011
  5. Aug 27, 2011 #4

    PeterO

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    I noticed you listed P = F x v as an appropriate formula. That is used for a body being driven at constant speed. This car is accelerating.

    I would be using P = work/time
    = (Force x distance)/time.
    = (mass x acceleration x distance)/time

    From the data given, you can calculate the acceleration, ad the time taken to get to the top of the hill - so away you go.
     
  6. Aug 27, 2011 #5

    PeterO

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    by the way - does the gradient 1 in 3 mean a rise of 1 metre for each 3 metres along the hill or does it mean a vertical rise of 1m for each 3m horizontal displacement.
    To find the angle involved you use sin for one, but tan for the other???
     
  7. Aug 27, 2011 #6

    v=22m/s
    s=100m

    acceleration
    v^2 = u^2 + 2as
    484=0 + 200a
    a=2.42m/s/s

    time
    t=v/a
    t=22/2.42
    t=9.09s

    Are these correct?


    Think its notation for vertical rise of 1m for each 3m horizontal displacement.
     
  8. Aug 27, 2011 #7

    PeterO

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    2.42 and 9.09 look just fine.

    Now don't forget the drive from the engine will be partially cancelled by the component of weight down the slope, so the engine may be pushing harder that you thought. That was why I queried what the "1 in 3" slope actually meant.
     
  9. Aug 27, 2011 #8
    So whats the next step?
    I have tried using 1/2mv^2 + mgh but I kept getting wrong answer.


    Maybe the 600kg is incorrect, can this question be solved without knowing weight of the car?
     
    Last edited: Aug 27, 2011
  10. Aug 27, 2011 #9

    PeterO

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    Of course now you know the time, you could just consider the gain in kinetic energy [moving fast now] plus the gain in Potential Energy [higher up].

    Power is (work)/(time) = (gain in energy)/(time)
     
  11. Aug 27, 2011 #10
    I've tried that I got incorrect answer of 70k W, should be 36.2W

    Maybe the 600kg I've quoted from section B is misleading. Can this question be solved without knowing the mass of the car?
     
  12. Aug 27, 2011 #11

    PeterO

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    what gain in altitude do you get, and what vale of g did you use?

    I get the 36.2 kW when I use the mass you said and the time you calculated?????
     
    Last edited: Aug 27, 2011
  13. Aug 28, 2011 #12
    h=100Cos(18.43)=94.87m
    g=9.8m/s/s



    Could you please solve this question for me, so I know where I am making the mistake?
     
  14. Aug 28, 2011 #13

    PeterO

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    THINK!!

    Do you really believe 100m of hill will rise 94.87 m when the gradient is 1 in 3???

    a gradient of 1 in 3 neans you go up about 1/3 the distance you move.

    That is where your problem lies.

    The cosine function is the wrong function!!
     
  15. Aug 28, 2011 #14
    Thanks that helps !

    I am still getting 36.4W not 36.2W :(

    KE = 0.5x600x22^2 = 145200J
    PE= 600x9.81x100Sin(18.43) = 186131J
    WD = 331331J

    Power = 331331/9.09 = 36450W
     
  16. Aug 28, 2011 #15

    PeterO

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    Could be rounding errors.

    I notice a lot of people on this forum use 9.81 for g. In our school system we always used 9.8

    I am very comfortable with the answer.

    by the way with g = 9.8, and the time being 100/11 rather than 9.09, it is 36.42 kW are you sure you were reading the "official" answer correctly, and weren't over looking the 4 ?
     
  17. Aug 28, 2011 #16

    gneill

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    These calculations look like they are for average power (total work divided by total time) rather than the minimum required power. Shouldn't the engine be developing more power at the 100m mark (where the velocity is greatest) in order to maintain the force required to keep the acceleration constant (P = F*V)?

    Is the question really asking for the minimum power?
     
  18. Aug 28, 2011 #17

    gneill

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    Heh. Problems in beginning physics classes often say to take g as 10 m/s2. Presumably this makes the numbers easier to deal with so that the student can concentrate on the physics rather than calculator operation. Later they find that 9.8 gives more realistic results. Still later, when they start worrying about error analysis and preserving significant figures through lengthier calculations, they start using 9.81, 9.807, or even 9.80665.

    With the advent of spreadsheets and computer algebra systems, it's convenient to assign g the most accurate value you have and then forget about it!
     
  19. Aug 28, 2011 #18

    PeterO

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    In the real world, you would have to allow for friction - which increases with speed.

    Since the car has reached 22 m/s [approx 80 km/h or 50 mph - not sure if you are from a metric country] air resistance for one would be becoming significant.

    We tend to ignore friction - or pay lip service by fixing it as XX-Newtons throughout.

    The reason we say minimum, is that it would be quite possible [easy in fact] to do this if the car had a 50 kW motor, or a 300 kW motor etc, but there is no way you could do it with a 30 kW motor..

    midnight here - won't be answering for a while.
     
  20. Aug 28, 2011 #19

    gneill

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    Even discounting all forms of friction, at the 100m mark the car must be developing power Force x Velocity at that instant. That'll be twice the average power over the whole run.
     
  21. Aug 28, 2011 #20
    Answer says 36.2kW however the sheet also says that some of the answer may be mis-typed :)
    Yes, the question is asking for the minimum power of the engine.

    Do you know if there's a way of solving it without knowing mass of the car?
     
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