## 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.

1. Michele_Laino

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

2. anonymous

ooh okie! :)

3. anonymous

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

4. Michele_Laino

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

5. 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

6. anonymous

ooh okie! so we get 478,237.5!!

7. Michele_Laino

ok! correct!

8. anonymous

yay!

9. anonymous

what happens next? :)

10. 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|

11. anonymous

ok! so we get 178,542 for PE!

12. Michele_Laino

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

13. anonymous

ooh okie! so KE = 299,695.5 !

14. Michele_Laino

correct! :)

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. 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|

19. 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?

20. Michele_Laino

yes at the top of second hill

21. anonymous

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

22. Michele_Laino

ok! correct!

23. Michele_Laino

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

24. anonymous

ok! so we get 298,756.25 !

25. Michele_Laino

that's right!

26. anonymous

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

27. Michele_Laino

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

28. anonymous

ohh okay!! where is position D?

29. Michele_Laino

|dw:1434906532984:dw|

30. anonymous

would that be bottom of second hill? oh ok!

31. Michele_Laino

ok! correct!

32. Michele_Laino

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

33. anonymous

34. Michele_Laino

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

35. Michele_Laino

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

36. anonymous

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

we have finished!

38. anonymous

ooh yay!! thank you!!:D

39. 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

40. Michele_Laino

:)

41. anonymous

oh okie!:)