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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.
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here is the drawing of your exercise: dw:1434905701577:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is all that is needed for part 1? :O

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, the total mechanical energy is: \[\large E = Mgh = 650 \times 9.81 \times 75 = ...Joules\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please 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
 one year ago
Best ResponseYou've already chosen the best response.0ooh okie! so we get 478,237.5!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what happens next? :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1when 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
 one year ago
Best ResponseYou've already chosen the best response.0ok! so we get 178,542 for PE!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! correct! so the kinetic energy KE is: \[\Large KE = 478237.5  178542 = ...Joules\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh okie! so KE = 299,695.5 !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh yay!! so this problem is complete? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait no or is that just from the bottom of the hill? it asks for top and bottom, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what we just calculated is for the bottom of the hill?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Now 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
 one year ago
Best ResponseYou've already chosen the best response.0wait, 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
 one year ago
Best ResponseYou've already chosen the best response.1yes at the top of second hill

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok! so for this top of second hill, we get 179,481.25 ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so the potential energy PE is: \[\Large PE = 478237.5  179481.25 = ...Joules\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! so we get 298,756.25 !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay!! so is this problem complete now? or is there more?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1there is more. at final position, (position D), the Potential energy of the crate is zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay!! where is position D?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434906532984:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would that be bottom of second hill? oh ok!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then the Kinetic Energy of our crate at position D, is: \[\Large KE = 478237.5Joules\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! what about potential energy?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is equal to the potential energy of the crate at position A or starting position dw:1434906675231:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Potential Energy at position D is zero since the elevation of the crate is h=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok! do we need to calculate anything for position A? or is that not needed since it is the starting position?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have finished!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh yay!! thank you!!:D

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1more precisely, since at position A the crate is at the rest, then at position A the Kinetic energy of our crate is zero
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