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Determining speed and height on a roller coaster

  1. Mar 27, 2015 #1
    1. The problem statement, all variables and given/known data
    One car (m=80.0 kg) tracks through the roller coaster in the following diagram. As it passes point A, it has a speed of 50.0 cm/s.

    a) Determine the speed at points B, C, and E
    b) If the speed at point D is _______, determine it's height (ignore friction)

    Ch 5 Roller Coaster.jpeg

    2. Relevant equations
    PE1 + KE1=PE2 + KE2
    PE=mgy
    KE=1/2mv^2

    3. The attempt at a solution
    a) I correctly got the velocity for point B as 44.3 m/s (1/2mv^2=mgy, 1/2(80 kg)v^2=(80kg)(9.80m/s^2)(100m). However, when I apply the same formula to points C and E my answer doesn't match that of the answer key.

    Point C: (80 kg)(9.80 m/s^2)(38 m)=1/2(80 kg)v^2, v=27.3 m/s. Answer key answer: 34.9 m/s

    Point E: (80 kg)(9.80 m/s^s)(20 m)=1/2(80 kg)v^2, v=19.8 m/s. Answer key answer: 39.6 m/s

    b) 1/2(80 kg)v^2=(80 kg)(9.80 m/s^2)y. I can simplify this equation but I don't know how to solve this as I would have 2 variables that I need to solve for.

    Thank you for your wonderful brains on this!
     
  2. jcsd
  3. Mar 27, 2015 #2

    jbriggs444

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    You quoted the equation: PE1 + KE1 = PE2 + KE2
    You used the equation: PE = KE
     
  4. Mar 27, 2015 #3

    Matternot

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    Don't get confused with ΔPE=ΔKE
     
  5. Mar 27, 2015 #4
    PE1 + KE1=PE2 + KE2 becomes PE=KE because, for example, at point C KE1=0 and PE2=0, so I end up using the PE1=KE2 equation.

    How am I confusing ΔPE with ΔKE?
     
  6. Mar 27, 2015 #5

    Matternot

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    KE1≠0 PE2≠0

    You haven't used the fact that it's moving at 0.5ms-1 at A

    You aren't confusing ΔKE with ΔPE. From your work it appears you are confusing ΔKE=ΔPE with KE=PE
     
    Last edited: Mar 27, 2015
  7. Mar 27, 2015 #6

    Matternot

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    Total energy conserved:

    mgy1+1/2 mv12=mgy2+1/2 mv22

    For point B:
    1/2(80kg)v2+(80kg)g0=(80kg)(9.80ms-2)(100m)+1/2(80kg)(0.5ms-1)2

    The only reason you got the first part "right" is because 1/2(80kg)(0.5ms-1) is really small compared to (80kg)(9.80ms-2)(100m) and so it didn't make a noticable difference to the answer
     
  8. Mar 27, 2015 #7

    Matternot

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    You want to use:

    PEA+KEA=(80kg)(9.80ms-2)(100m)+1/2(80kg)(0.5ms-1)2=PEx+KEx=mgy+1/2mv2

    For all but the last parts, y is known

    For the last part, v is known
     
    Last edited: Mar 27, 2015
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