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Average power question

  1. Dec 10, 2007 #1
    The question reads as such:

    A car of mass 1500kg traveling at 24 m/s is at the foot of a hill that rises 120 m in 2.0 km. At the top of the hill the speed of the car is 10 m/s. Find the average power delivered by the car's engine, neglecting any frictional losses.

    The book says the answer is 18 kW. My work gets me the answer 11980.8 W, my friend got 15 kW.

    Here's what I did:

    The actual hill is 2003.6 m long (Pythagorean).

    The angle of incline is 3.44 deg (sin^-1(120/2003.6)).

    Acceleration due to gravity is -9.8 sin(3.44), or -.588 m/s^2.

    The car decelerates from 24 to 10 m/s, so at = -14.

    d = 1/2 at^2 + vi*t

    Sub in at, d = 1/2 (-14)t + vi*t This is the step that might be wrong, I suppose, because this idea just came to me out of the blue. It SHOULD work, since a*t is a*t regardless, but meh, I've been wrong before.

    2003.6 = .5(-14)t + (24)t

    t = 117.9

    at = -14

    a = -14/117.9 = -.118

    a(gravity) + a(engine) = a(total)

    -.588 + a(engine) = -.118

    a(engine) = .47 m/s^2

    f(engine) = ma

    f(engine) = 1500 * .47 = 705 N

    p(engine) = (f*d)/t

    p(engine) = (705 * 2003.6)/117.9 = 11980.8 W

    Any insight into this problem?
  2. jcsd
  3. Dec 10, 2007 #2

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    With the infmn you have provided, the problem can't be solved. Try to write down the complete problem, exactly as it appears. For example, you have mentioned the angle of inclination, but not at the beginning.
  4. Dec 10, 2007 #3
    Could someone at least tell me if it's permissible to do the at sub in at step 6?
  5. Dec 10, 2007 #4
    And that IS the full problem, want a picture of it? I typed it word for word. I found out the angle using sin(theta) = opposite/hypotenuse.
  6. Dec 10, 2007 #5
  7. Dec 10, 2007 #6


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    I don't see anything wrong with your steps so far, but maybe you should try another approach and see if you get the same answer.

    1. Find the difference between total energy (potential + kinetic) of the car at the bottom and top of the hill. This difference must be accounted for by the car's engine (conservation of energy).

    2. Find the acceleration of the car using kinematics (i.e. Vf^2 = Vi^2 + 2ad)

    3. Find the time to travel from bottom to top using kinematics (i.e. Vf = Vi + at)

    4. Calculate average power assuming constant acceleration using P = E/t.

    I hope this helps.
  8. Dec 11, 2007 #7

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    I'm very sorry I missed out on some of the details when reading your problem. The answer is 18028.5 W = 18 KW. Your method is good.

    There's no need to take the angle of incline. Elevation of 2 km in 120 m means it's 2 km along the road.
  9. Dec 11, 2007 #8

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    Just outlining your own method in short. I think you plugged in the value of (-at) instead of +at.

    E = Energy expended = Ef –Ei = mgh + ½ mvf^2 -1/2 mvi^2.

    v = u-at => at = 24-10 = 14.

    d = ut – ½ at^2 = 24t – ½ * 14*t => t = d/17, where d = 2000m.

    P = E/t = 18000, after simplification.
  10. Dec 11, 2007 #9
    I tried the 2000 m way as well and the answer wasn't significantly different (the time was .3 seconds different and the angle was the same I think; the final answer was almost exactly the same). I was a little confused about if the 2000 m was along the road or like geographical though, thanks for clarifying.

    I don't know if I did the -at thing, the acceleration IS negative so that should make at negative, no?

    I'll do it your way though, danke =)
  11. Dec 11, 2007 #10

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    Yes, I checked your work. About the 'at', there's no problem.

    Anyway, for this type of problem, use conservation of energy.
  12. Dec 11, 2007 #11
    Ah, okay. I talked to my teacher about this today. Using conservation of energy you still get 11960 (close enough to my previous answer, considering the number of times I rounded in that method). The book's answer was incorrect, and we even figured out why:

    E = Energy Expended = Ef - Ei = mgh + 1/2 mvf^2 - 1/2 mvi^2, just as you said. E = 1407000 joules, whereas your E value was calculated at 2121000 joules. You made the same mistake the bookmakers did; you used 24 m/s as your vf and 10 m/s as your vi. If that were the case then yes, the engine would have an average power of 18 kW, but it's the other way around.

    You're totally right about using conservation of energy though, it's much faster (although apparently a lot easier to mix up).
  13. Dec 11, 2007 #12

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    I did not.

    I used 24 m/s for u, by which I meant vi, and 10 m/s for v, by which I meant vf. It's there in my last but one post.

    Anyway, I'll go through it later. No time now. The method is clear. BTW, what's the correct answer?
  14. Dec 11, 2007 #13
    It has to be 11960 W (11.96 kW). I didn't quite get the "u" part when I read through your post, maybe it's some notation I'm unfamiliar with? But just the fact that you have "at = 24-10 = 14." shows that you have vf and vi mixed up in at least one place. at = -14, not +14. The car is slowing DOWN. This is what I was talking about by saying "You made the same mistake the bookmakers did; you used 24 m/s as your vf and 10 m/s as your vi. If that were the case then yes, the engine would have an average power of 18 kW, but it's the other way around."

    You have at = 14, meaning the change in velocity is +14 m/s. If that's the case then vi = 10 m/s and vf = 24 m/s and the engine indeed has to work harder. The problem is the other way around though.

    When you look back, tell me what you calculate E to be? Show all of the numbers and where they go while doing it too, if you would =)
  15. Dec 11, 2007 #14

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    I have written "v = u-at => at = 24-10 = 14." This means u=vi and v=vf. You see, I've already assumed that the car is slowing down, and that's why there's a minus sign in the eqn. So, I'm just finding the magnitude of a, and 'a' is actually the deceleration. That has absolutely no significance on the whole treatment.

    I'll write down all the steps for your satisfaction, when I get time. I'm not saying that I haven't made arithmetical mistakes, but the at=14 is not one of them.

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