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Potential Energy Car Downhill Question

  1. Nov 17, 2005 #1
    I was set this question and have no clue how to solve it...

    A 1100 kg car racing up a mountain road runs out of gas at a height of 36 m while traveling at 16 m/s. Cleverly, the driver shifts into neutral and coasts onward. Not having any brakes, at what speed will the car reach the bottom of the hill?

    I found mgh of the car as 1100 x 9.81 x 36 and then equated this to 1/2mv2 to try and find velocity and I got 26.5m/s but this is wrong...any help? Thanks
  2. jcsd
  3. Nov 17, 2005 #2


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    what about the kenetic energy of the car when it is at 36m?
  4. Nov 17, 2005 #3
    PE of car at 36m = mgh = 1100 x 9.81 x 36 = 388476
    PE of car at 0m = mgh = 0

    KE of car at 36m = 1/2mv2 = 1/2 x 1100 x v2 = 550v2
    KE of car at 0m = 1/2mv2 = 550v2

    Now where can I go from here?
  5. Nov 17, 2005 #4


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    what about Energie(s) intial = Energie(s) final
  6. Nov 17, 2005 #5
    OK here's what I got for initial energies = final energies...(v = at the peak of the mountain and V = at the bottom)

    388476 + 550v2 = 0 + 550V2

    623.27 + 550v = 550V

    1.133 + v = V

    V = 1.133 + 16 = 17.133

    But this is wrong...
  7. Nov 17, 2005 #6
    The total mechanical energy of the car is the sum of its kinetic energy and its potential energy, both of which you know for the height of 36 m.

    By the time the car reaches the ground, all of its mechanical energy (which is conserved) is kinetic energy - or, in other words, all of its original potential energy has been converted to kinetic energy (in addition to its original kinetic energy).
  8. Nov 17, 2005 #7
    What exactly did you do here? It looks like you took square roots of some of the numbers separately, which really isn't allowed. The first equation is correct and you can solve for V simply by diving both sides with 550kg and then taking the square root. You should note however, that all the numbers must be within the same square root, you can't take the root from each one separately.
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