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Using power to find velocity (a car meets a hill)

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    A car encounters an inclined plane

    Given- Weight of car in N (6500), velocity on the flat surface (22.5 m/s), power of the engine (78000), incline of the hill (8.1 degrees)

    Want to find- velocity on the hill (power and restistive forces remain constant)

    2. Relevant equations

    P = Fvcos(theta)

    3. The attempt at a solution

    I found F by plugging in the power and velocity. I then subtracted the gravitational force due to the hill from this number to get the resultant force. plugged this in to P = Fvcos(theta) and got a number bigger than the original...

    urgent help would be greatly appreciated since HW is due in 8 minutes, but i'm more worried about the concept for the test tomorrow. thanks.
  2. jcsd
  3. Oct 12, 2008 #2


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    Staff: Mentor

    The engine power is constant so the car is traveling at constant speed so that resistance is constant.

    When going up hill - the car starts increasing its gravitational potential energy - and if the car's power output is constant, then the car's kinetic energy must be decreasing.
  4. Oct 12, 2008 #3
    okay, so if i have to use kinetic energy, does that need i need to find the mass of the car (1/2mv^2)? that doesnt seem right...
  5. Oct 12, 2008 #4


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    Science Advisor

    edit: Ok I think I have it this time

    Your flat ground situation is just P = FV. Move it around to get F = P/V. This is the force coming from the engine; it does not change. When you get on the hill, gravity applies a force against the motor as Wsin(theta). With this new net force, you find the new velocity.

    Flat ground:
    P = FV (start with this)
    F = P/V (solve for force)

    P = (F + gravity)V

    P is the same, F you find out, gravity is Wsin(theta), V is your answer. They are ADDED together because F and gravity represent DRAG as opposed to the force you are applying.
    Last edited: Oct 12, 2008
  6. Oct 12, 2008 #5
    i tried this also, maybe i made a calculation error...
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