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

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- Michele_Laino

here is the drawing of your exercise:
|dw:1434905701577:dw|

- anonymous

ooh okie! :)

- anonymous

that is all that is needed for part 1? :O

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## More answers

- Michele_Laino

now, the total mechanical energy is:
\[\large E = Mgh = 650 \times 9.81 \times 75 = ...Joules\]

- 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

ooh okie! so we get 478,237.5!!

- Michele_Laino

ok! correct!

- anonymous

yay!

- anonymous

what happens next? :)

- 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

ok! so we get 178,542 for PE!

- Michele_Laino

ok! correct! so the kinetic energy KE is:
\[\Large KE = 478237.5 - 178542 = ...Joules\]

- anonymous

ooh okie! so KE = 299,695.5 !

- Michele_Laino

correct! :)

- anonymous

ooh yay!! so this problem is complete? :O

- anonymous

oh wait no or is that just from the bottom of the hill? it asks for top and bottom, right?

- anonymous

what we just calculated is for the bottom of the hill?

- 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

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

yes at the top of second hill

- anonymous

oh ok! so for this top of second hill, we get 179,481.25 ?

- Michele_Laino

ok! correct!

- Michele_Laino

so the potential energy PE is:
\[\Large PE = 478237.5 - 179481.25 = ...Joules\]

- anonymous

ok! so we get 298,756.25 !

- Michele_Laino

that's right!

- anonymous

yay!! so is this problem complete now? or is there more?

- Michele_Laino

there is more.
at final position, (position D), the Potential energy of the crate is zero

- anonymous

ohh okay!! where is position D?

- Michele_Laino

|dw:1434906532984:dw|

- anonymous

would that be bottom of second hill? oh ok!

- Michele_Laino

ok! correct!

- Michele_Laino

then the Kinetic Energy of our crate at position D, is:
\[\Large KE = 478237.5Joules\]

- anonymous

ok! what about potential energy?

- Michele_Laino

it is equal to the potential energy of the crate at position A or starting position
|dw:1434906675231:dw|

- Michele_Laino

Potential Energy at position D is zero since the elevation of the crate is h=0

- 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

we have finished!

- anonymous

ooh yay!! thank you!!:D

- 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

:)

- anonymous

oh okie!:)

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