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Conservation of energy of a roller coaster car problem

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A roller coaster car starts from rest and rolls down a frictionless track, reaching Vf ( final speed) at the bottom.

    if you want the car to go two times as fast at the bottom by what fator must you increase the height of the track

    2. Relevant equations

    3. The attempt at a solution
    So according to the conservation of energy
    KEi+PEi = KEf+PEf

    in this case the KEi = 0 and PEf = 0
    PEi = KEf
    mgh= 1/2mv^2
    mass cancels

    so v= square root of 2gh

    after this i have though to set the equations like this

    vf= sqaure root of 2gh
    2vf= square root of 2gh

    that what i have so far but i dont know how to continue on
  2. jcsd
  3. Nov 8, 2008 #2
    You're almost there: you have

    vf=square root of 2gh.

    You want a new height H, such that

    2vf =square root of 2gH,

    so what should H be?
  4. Nov 8, 2008 #3
    ok so this is what i did
    i set both equation to vf

    vf=square root of 2gh
    vf= (square root of 2gH)/2

    set them equal to each other and divided by 2

    2 (square root of 2gh) = suare root of 2gH

    (2*(square root of 2gh))^2 = gH

    divided by g

    ((2 *(square root of 2gh))^2)/ g = H


    4*2gh/g = H

    g's cancel


    that means that to increase vf by two we must increse the height by a factor of 8 right???
  5. Nov 8, 2008 #4
    No...but almost: you made a simple mistake. Try again. Can't you see the answer directly by looking at the equations?
  6. Nov 9, 2008 #5
    Ok then it has to increse by a factor of 4!
  7. Nov 9, 2008 #6
    There you go!
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