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Conservation of energy of a truck problem

  1. Oct 31, 2004 #1
    I tried many ways to perfomr this problem, but i just cant figure out whether or not Im using the right values.

    "A 25 ton truck loses it brakes and reaches the bottom of a hill with a speed of 60 mph.
    Fortunately, there is a runaway truck ramp which is inclined at angle of 15 degrees to the horizontal."
    Assuming no losses, what distance does the truck travel along the runaway truck ramp? (miles) ?

    The KE of the truckjust before it goes up the ramp is 6010000 ft-lb.

    What do I need to do to get the distance the truck travel along the ramp? The length of the ramp is 1 mile if that means anything.

    Heres a formula Ive used : KE1+PE1+Work=KE2+PE2+Eloss

    Im not sure what force i need to use. I would guess the 6010000 I found.
  2. jcsd
  3. Oct 31, 2004 #2


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    Draw a picture of the hill.
    You know how high it will go by using conservation of energy.
    At the bottom it will only have kinetic energy (which value you know).
    At the highest point it will only have grav. potential energy (which must equal the total energy).
  4. Oct 31, 2004 #3

    Doc Al

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    Staff: Mentor

  5. Oct 31, 2004 #4
    you can delete this post
  6. Oct 31, 2004 #5
    First off, I don't get 6010000 ft*lbs for the Kinetic Energy of the truck, Kinetic energy is measured witht the equation (1/2 m * v^2) where, in the english system, m must be in slugs (lbs/g), and v in (ft/s).

    You should look at the type of energy conversion you are doing, you start with a force which is entirely kinetic energy, and you end with no kinetic energy, because your velocity is 0.

    Because of the Laws of Conservation of Energy, you cannot simply lose energy, so your energy must have been converted to another form. Because their is no friction acting, the energy must all be transfered to potential energy.

    The equation for potential energy is (mgh) where m equals the mass of the truck, g is the acceleration due to gravity (32.2 ft/s) and h is the height above the starting height (the bottom of the hill)

    Because these two energies must be equal you can set 1/2mv^2 = mgh and solve for h, then use trigonometry to solve for the distance along the ramp the truck must travel to get to the necessary height.

    Hope this helps,

  7. Oct 31, 2004 #6
    6010000 ft*lbs KE i got from dividing 8200000J by 1.36J. They just gave me some conversion factors to work with. Im thinking i should use F*Scos(theta) which equals work.
  8. Oct 31, 2004 #7
    I converted my initial KE from ft-lbs to J and had that = to my inital KE. So i converted 6010000 ft-lb to J by multiplying by 1.36J/ft-lb which = to 8200000

    I use the equation KE1+PE1+Work=KE2+PE2+Eloss
    Since no energy was loss I cancelled that out. I cancelled out PE1 because the height was 0. So I ended up with KE1+Work=KE2+PE2. Since the ramp is 1 mile long, I converted that to meters by multiplying 1mile x 1609.344m/mile which is just 1609.344.

    The values are now
    I came out with s=16000 m
    I converted that to miles by dividing by 1609.344 m to get around 9.94 miles for my answer but after doing all that, I go the wrong answer.
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