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Amusement park ride problem

  1. Sep 16, 2003 #1
    Hi Everyone,

    Thank you for dduardo and HllsofIvy for helping me on my last question. Once again, I am having problems with my physics homework. Here is my question:

    In an amusement park ride, people in a car are dropped some horrible distance, screaming for their lives. The car rounds a curved track and is brought to a halt be a braking force on a horizontal piece of track. The car a=gas a mass if 1100kg. Point A is 50m off the ground. Point B is 20m off the ground. And point C is 5m off the ground. The track is frictionless until you get to the horizontal braking section. The braking force is 2500lb. 1.) Find the velocity of the car at B and when it reaches the horizontal track. 2.)How far will the car move on the horizontal track? 3.)Then re-work the problem with the braking force kicking in at point B rather than at the horizontal track.

    Here is my work, which is more than likely wrong because I have no ideal how what a breaking force is or how to calculate that into my problem:

    W = m * g * y
    W = 1100kg * 9.8 N * 30 m = 323,400J

    W = F * d
    323,400J/30m= 10780kg (m/s^2)

    a = F/m = 10780kg m/s^2 /1100kg =9.8 m/s^2

    v^2=v0^2 +2 average acceleration (x – x0 )
    v^2 =0 + 2 (9.8 m/s^2) (30m-0m)
    square root of v^2 = square root of (508 m/s)^2
    1.) Therefore the velocity at B= 22.5 m/s

    W = mv^2 /2 = 1100kg(22.5m/s)^2 /2 = 278,437.5J

    2.) W = m * a * d = 278,437.5J * 9.8m/s^2 * d = 25.83m was the distance

    3.) I have no ideal how to work the third part.
  2. jcsd
  3. Sep 16, 2003 #2


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    Science Advisor

    Looks like a good start: you calculate the potential energy of the car at the start. But why find the force and acceleration? You aren't asked that. You are asked the speed at each of several points. You can find that by "conservation of energy": find the potential energy at points B and C (is that the beginning of the "horizontal" part? You don't say.) Since energy is kinetic energy + potential energy, you must have (1/2)mv^2+ potential energy= potential energy at A or (1/2)m v^2= potential energy at A- potential energy at the new point (B or C). That allows you to calculate the velocity at each point.

    Now look at what happens on the horizontal part. There is some "braking" force F= 2500 lb (lb?? POUNDS?? but everything else is in metric. Are you sure it wasn't 2500 N?) and the work done by that is 2500x where x is the distance on the horizontal part. Again, find the potential energy at the end (the kinetic energy is 0 since the car is stopped) and add: 2500x+ potential energy= total energy (which is the same as the same as before).
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