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anonymous

  • one year ago

4. An 800.0 kg roller coaster car is at rest at the top of a 95 m hill. It rolls down the first drop to a height of 31 m. When it travels to the top of the second hill, it is moving at 28 m/s. It then rolls down the second hill until it is at ground level. Draw a picture of the roller coaster drop and calculate the kinetic and potential energy at the top and bottom of each hill.

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  1. anonymous
    • one year ago
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    Can you help me with this?

  2. nikato
    • one year ago
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    |dw:1433303891314:dw| so this would be your diagram. ok?

  3. anonymous
    • one year ago
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    So how do I do the rest? @nikato

  4. nikato
    • one year ago
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    using the formula for potential energy, you can find the PE at point A remember PE= mgh

  5. nikato
    • one year ago
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    so 800(10)(95)= 760000 is the potential and since it's at rest kinetic is 0 this is all at point A

  6. nikato
    • one year ago
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    find PE at point B 800(10)(31)= 248000 to find kinetic, subtract this from the potential at point A because i'm assuming that energy is conserved so 760000-248000= 512000 for kinetic at B

  7. nikato
    • one year ago
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    for C, we are given the velocity so we can find the kinetic energy \[KE= \frac{ 1 }{ 2 }mv ^{2}\]

  8. nikato
    • one year ago
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    1/2 (800)(28)^2= 313600 again subtract this from PE at A to find potential at C since energy is conserved so 760000-313600= 446400

  9. nikato
    • one year ago
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    assuming that it continues on ground level, all the potential energy at A will be kinetic at D if energy is conserved. so K at D= 760000 and P=0 since it's at ground level at h=0 remember all my calculations are assuming that you want gravity to be 10 m/s^2

  10. abb0t
    • one year ago
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    @Jhannybean

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