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Roller Coaster-Gravitational potential

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data

    The figure shows a roller coaster car mass m= 3,000kg, on a portion of a roller coaster ride. The height difference between points A and B is 35m. The height difference between B and C is 23m. The car starts from rest at point A. Take y=0 point to be point B. For parts a.,b., and c., assume that the roller coaster track is frictionless.



    2. Relevant equations
    a.) Compute the gravitational potential energy of the car at points A, B, and C.
    b.) Compute the kinetic energy of the car at point B. Compute its speed there. What physical principle did you use to do this calculation?
    c.) compute the kinetic energy and speed of the car at point C.
    d.) Parts a., b., and c. assume that the track is frictionless. However, the measured speed of the car at point B is found to be V=20m/s. This is less than the speed that you should have computed in part b. This means that friction cannot be neglected. In this case, how much work is done by friction when the car moves from point A to point B? What physical principle did you use to compute this work?



    3. The attempt at a solution

    a.) gravitational potential energy =m*g*h

    Point A= 3000*9.8*35= 1.029E+6J

    Point B= 3000*9.8*0= 0J

    Point C= 3000*9.8*23= 676,200J

    b.) E=(1/2)m*v^2+m*g*h+(1/2)kx^2

    At point A, there is no kinetic energy and no elastic energy
    At point B, there is no gravitational potential energy or elastic energy
    And since Ea=Eb
    m*g*h=(1/2)m*v^2

    Vb=26.19 m/s

    KE=(1/2)m*v^2
    KE= 1.029E+6J, same as the gravitational potential energy at point A.

    C.) There is gravitational potential energy and kinetic energy at point C, so:

    (1/2)m*v^2=(1/2)m*v^2+m*g*h

    Vc= 15.34 m/s

    KE= 352,973 J

    d.) I wanted to make sure I was doing everything right before I go on lol. Can anyone verify my answers thus far?
     
  2. jcsd
  3. Mar 21, 2009 #2
    Everything seems correct to me. At least, the equations are.. :wink:
    Didn't do the calculations but I reckon you're right.
     
  4. Mar 21, 2009 #3
    Okay thanks! Could you possibly push me in the right direction on how to approach part d?
     
  5. Mar 21, 2009 #4

    arildno

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    d) Hint: What does the work-energy theorem say?
     
  6. Mar 22, 2009 #5
    Work-energy theorem:

    W=f*d*cos(theta)

    and

    W= KE final- KE initial

    so:

    (1/2)*3000*(26.19)^2- (1/2)* 3000*(20)^2 = 428,874J

    Is that right?
     
  7. Mar 25, 2009 #6
    anyone?
     
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