anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
here is the drawing of your exercise: |dw:1434905701577:dw|
anonymous
  • anonymous
ooh okie! :)
anonymous
  • anonymous
that is all that is needed for part 1? :O

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More answers

Michele_Laino
  • Michele_Laino
now, the total mechanical energy is: \[\large E = Mgh = 650 \times 9.81 \times 75 = ...Joules\]
Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
ooh okie! so we get 478,237.5!!
Michele_Laino
  • Michele_Laino
ok! correct!
anonymous
  • anonymous
yay!
anonymous
  • anonymous
what happens next? :)
Michele_Laino
  • Michele_Laino
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|
anonymous
  • anonymous
ok! so we get 178,542 for PE!
Michele_Laino
  • Michele_Laino
ok! correct! so the kinetic energy KE is: \[\Large KE = 478237.5 - 178542 = ...Joules\]
anonymous
  • anonymous
ooh okie! so KE = 299,695.5 !
Michele_Laino
  • Michele_Laino
correct! :)
anonymous
  • anonymous
ooh yay!! so this problem is complete? :O
anonymous
  • anonymous
oh wait no or is that just from the bottom of the hill? it asks for top and bottom, right?
anonymous
  • anonymous
what we just calculated is for the bottom of the hill?
Michele_Laino
  • Michele_Laino
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|
anonymous
  • anonymous
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?
Michele_Laino
  • Michele_Laino
yes at the top of second hill
anonymous
  • anonymous
oh ok! so for this top of second hill, we get 179,481.25 ?
Michele_Laino
  • Michele_Laino
ok! correct!
Michele_Laino
  • Michele_Laino
so the potential energy PE is: \[\Large PE = 478237.5 - 179481.25 = ...Joules\]
anonymous
  • anonymous
ok! so we get 298,756.25 !
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
yay!! so is this problem complete now? or is there more?
Michele_Laino
  • Michele_Laino
there is more. at final position, (position D), the Potential energy of the crate is zero
anonymous
  • anonymous
ohh okay!! where is position D?
Michele_Laino
  • Michele_Laino
|dw:1434906532984:dw|
anonymous
  • anonymous
would that be bottom of second hill? oh ok!
Michele_Laino
  • Michele_Laino
ok! correct!
Michele_Laino
  • Michele_Laino
then the Kinetic Energy of our crate at position D, is: \[\Large KE = 478237.5Joules\]
anonymous
  • anonymous
ok! what about potential energy?
Michele_Laino
  • Michele_Laino
it is equal to the potential energy of the crate at position A or starting position |dw:1434906675231:dw|
Michele_Laino
  • Michele_Laino
Potential Energy at position D is zero since the elevation of the crate is h=0
anonymous
  • anonymous
oh ok! do we need to calculate anything for position A? or is that not needed since it is the starting position?
Michele_Laino
  • Michele_Laino
we have finished!
anonymous
  • anonymous
ooh yay!! thank you!!:D
Michele_Laino
  • Michele_Laino
more precisely, since at position A the crate is at the rest, then at position A the Kinetic energy of our crate is zero
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
oh okie!:)

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