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Potential Energy of a roller coaster cart

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A 900 kg roller-coaster car is initially at the top of a rise, at point A. It then moves 165 ft, at an angle of 40.0° below the horizontal, to a lower point B.

    Choose the car at point B to be the zero configuration for gravitational potential energy of the roller coaster-Earth system. Find the potential energy of the system when the car is at points A and B, and the change in potential energy as the coaster moves.


    2. Relevant equations

    U=mgy


    3. The attempt at a solution

    I've found that the potential energy at point B is 0 (obviously), but I can't find it at A.

    I'm pretty sure it's U=900(9.8)h

    But I'm not sure how to find the height. Wouldn't it be something like 165sin40?

    Edit: D'oh. Figured it out. the distance is in feet, gotta convert it to meters. Stupid conversions. :P
     
    Last edited: Mar 11, 2009
  2. jcsd
  3. Mar 12, 2009 #2

    CompuChip

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    Science Advisor
    Homework Helper

    Two approaches to avoid this kind of errors:

    1) First write down all the data in a list, including units. Before you even start writing down formulas, convert everything to SI units (m, s, kg, J, W, ....).

    m = 900 kg
    g = 9.81 m / s^2
    h = ... ft [tex]\widehat{=}[/tex] ... m
    where I used [tex]\widehat{=}[/tex] as a symbol for "corresponds to" (people tend to use just the equality sign, but I find it better to use this symbol because feet aren't meters, they can just be used to measure the same thing).

    2) Try dragging the units along in your calculation (for some people this makes things more complicated, but IMO once you get used to doing so it can catch a lot of errors for you).

    U = m g h = (900 kg) * (9.81 m/s^2) * (.... ft) = .... (kg m ft / s^2)

    Errr... kg m ft / s^2 is not Joules, because Joules are kg m^2 / s^2. There must be a conversion from ft to m somewhere.

    Advantage of this last approach is, that if it had given you kg m / s^2, for example, you would have known that you had made a calculation error. Because it should give kg m^2 / s^2, and there is no way you can get an extra length inside there, so you have forgotten to multiply by a length. (Only this does not work when rewriting formula's, and you doubt whether the 2pi has to be in the denominator or the numerator of the fraction, otherwise it will help you catch errors).
     
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