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anonymous

  • one year ago

A 650.0 kg roller coaster car is at rest at the top of a 75 m hill. It rolls down the first drop to a height of 28 m. When it travels to the top of the second hill, it is moving at 23.5 m/s. It then rolls down the second hill until it is at ground level. Part 1: Draw a picture of the roller coaster drop. Part 2: Calculate the kinetic and potential energy at the top and bottom of each hill.

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  1. Michele_Laino
    • one year ago
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    here is the drawing of your exercise: |dw:1434905701577:dw|

  2. anonymous
    • one year ago
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    ooh okie! :)

  3. anonymous
    • one year ago
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    that is all that is needed for part 1? :O

  4. Michele_Laino
    • one year ago
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    now, the total mechanical energy is: \[\large E = Mgh = 650 \times 9.81 \times 75 = ...Joules\]

  5. Michele_Laino
    • one year ago
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    please note that the mechanical energy of the crate when it is at other position has to be equal to that one, since the gravitational field is conservative

  6. anonymous
    • one year ago
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    ooh okie! so we get 478,237.5!!

  7. Michele_Laino
    • one year ago
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    ok! correct!

  8. anonymous
    • one year ago
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    yay!

  9. anonymous
    • one year ago
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    what happens next? :)

  10. Michele_Laino
    • one year ago
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    when the crate is at position B, then its potential energy PE is: \[\Large PE = Mgh = 650 \times 9.81 \times 28 = ...Joules\] |dw:1434905977803:dw|

  11. anonymous
    • one year ago
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    ok! so we get 178,542 for PE!

  12. Michele_Laino
    • one year ago
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    ok! correct! so the kinetic energy KE is: \[\Large KE = 478237.5 - 178542 = ...Joules\]

  13. anonymous
    • one year ago
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    ooh okie! so KE = 299,695.5 !

  14. Michele_Laino
    • one year ago
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    correct! :)

  15. anonymous
    • one year ago
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    ooh yay!! so this problem is complete? :O

  16. anonymous
    • one year ago
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    oh wait no or is that just from the bottom of the hill? it asks for top and bottom, right?

  17. anonymous
    • one year ago
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    what we just calculated is for the bottom of the hill?

  18. Michele_Laino
    • one year ago
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    Now at position C, the Kinetic Energy KE is: \[\large KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 650 \times {23.5^2} = ...Joules\] |dw:1434906217586:dw|

  19. anonymous
    • one year ago
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    wait, so just now, what we calculated, that is for the bottom of the hill? (point B) and now, point C we are calculating for the top of the hill?

  20. Michele_Laino
    • one year ago
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    yes at the top of second hill

  21. anonymous
    • one year ago
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    oh ok! so for this top of second hill, we get 179,481.25 ?

  22. Michele_Laino
    • one year ago
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    ok! correct!

  23. Michele_Laino
    • one year ago
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    so the potential energy PE is: \[\Large PE = 478237.5 - 179481.25 = ...Joules\]

  24. anonymous
    • one year ago
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    ok! so we get 298,756.25 !

  25. Michele_Laino
    • one year ago
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    that's right!

  26. anonymous
    • one year ago
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    yay!! so is this problem complete now? or is there more?

  27. Michele_Laino
    • one year ago
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    there is more. at final position, (position D), the Potential energy of the crate is zero

  28. anonymous
    • one year ago
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    ohh okay!! where is position D?

  29. Michele_Laino
    • one year ago
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    |dw:1434906532984:dw|

  30. anonymous
    • one year ago
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    would that be bottom of second hill? oh ok!

  31. Michele_Laino
    • one year ago
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    ok! correct!

  32. Michele_Laino
    • one year ago
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    then the Kinetic Energy of our crate at position D, is: \[\Large KE = 478237.5Joules\]

  33. anonymous
    • one year ago
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    ok! what about potential energy?

  34. Michele_Laino
    • one year ago
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    it is equal to the potential energy of the crate at position A or starting position |dw:1434906675231:dw|

  35. Michele_Laino
    • one year ago
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    Potential Energy at position D is zero since the elevation of the crate is h=0

  36. anonymous
    • one year ago
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    oh ok! do we need to calculate anything for position A? or is that not needed since it is the starting position?

  37. Michele_Laino
    • one year ago
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    we have finished!

  38. anonymous
    • one year ago
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    ooh yay!! thank you!!:D

  39. Michele_Laino
    • one year ago
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    more precisely, since at position A the crate is at the rest, then at position A the Kinetic energy of our crate is zero

  40. Michele_Laino
    • one year ago
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    :)

  41. anonymous
    • one year ago
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    oh okie!:)

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