Question

MEDAL!!!!


A 1000 kg rollercoaster is released from rest from point D on the rollercoaster track. Assume no friction and no air resistance. How fast is the rollercoaster moving at point F? (Point D is 75 m high ; point F is 35 m high)

Diagram attached below.

Answer 1

@jim_thompson5910

Answer 2

So would this be related to the Law of Conservation of Energy: (PE + KE) = (PE + KE)

Answer 3

I think once you get 392,000 J for the KE at point F, you use that to find the velocity v

Answer 4

so like:
392,000 = 1/2mv^2

Answer 5

yes

Answer 6

and then:
392,000 = 1/2(1000)v^2

Answer 7

392,000 = 500v^2
784 = v^2
v= 28 m/s

Answer 8

very good

Answer 9

so acceleration is
\[\frac{ \Delta V }{ time }\]

Answer 10

over change in time

Answer 11

i have to find the two velocities?

Answer 12

how would i find the time?

Answer 13

the first velocity is 0 since it starts at rest

Answer 14

I'm not sure about the time part

Answer 15

can i use:
distance = 1/2gt^2

Answer 16

maybe there's another formula tying together distance, velocity and acceleration

Answer 17

distance for point d = 1/2gt^2
75 m = 1/2(9.8m/s^2)t^2

Answer 18

I think that's for free fall only?

Answer 19

idk if that applies along the curve

Answer 20

do u know a formula that i can use to find the time?

Answer 21

no I don't sadly

Answer 22

do they give any other info?

Answer 23

no...

Answer 24

but can we do the second question?

Answer 25

I'm not sure. They want the acceleration at point F right?

Answer 26

yes

Answer 27

I'd get a second opinion from a physics expert because I'm not sure about the second part.

Answer 28

here is the second question:
2. a 400 kg rollercoaster is released from rest at point B on the rollercoaster track above. assume no friction and no air resistance. if it is moving at 20 m/s at point c ( 60 m high), how high is point B?

Answer 29

PE at point C

PE = m*g*h
PE = 400*9.8*60
PE = 235,200


KE at point C

KE = (1/2)*m*v^2
KE = (1/2)*400*20^2
KE = 80,000


Total Energy (TE) at point C

TE = PE + KE
TE = 235,200 + 80,000
TE = 315,200

-------------------------------------------------------

TE at point B

TE = PE + KE
TE = m*g*h + (1/2)*m*v^2
315,200 = 400*9.8*h + (1/2)*400*0^2

solve for h

Answer 30

For Q1, where does it ask for the acceleration?
the only acceleration is due to gravity, and you need more info to calculate it
(though at the top of a "hill" gravity is perpendicular to direction, and acceleration is 0)

Answer 31

i think i got.

since \[(PE + KE)_D = (PE + KE)_F\]
and we already know the potential energy, which is 735,000 joules and the kinetic energy would be 0 joules.
\[(735,000 + 0)_D = (343,000 + KE)_F\]

wouldn't we have to subtract 735,000 - 343,000 to get the kinetic energy?

Answer 32

are you referring to problem #1 or #2 ?

Answer 33

#1

Answer 34

yeah I agree with the work you wrote on the paper (in the attachment)

Answer 35

movement = kinetic energy (which doesn't necessarily have to mean acceleration)

Answer 36

then you did a bit more work to find that v = 28 m/s

Answer 37

yeah i just realized that :)

Answer 38

we can work on problem #2 now

Answer 39

I wrote out #2 above

Answer 40

oh okay i will look on that right after i write down my work on the paper.

here is another question:
a 600 kg rollercoaster is released from rest at point A (150m high). what is its potential energy at A? how fast is it going at point C (40 m high)? what is its kinetic energy at point C? what is its potential energy at point D?

Answer 41

`a 600 kg rollercoaster is released from rest at point A (150m high). what is its potential energy at A?`

use PE = m*g*h

m = 600 is the mass
g = 9.8 is the approximate acceleration of gravity
h = height

Answer 42

@jim_thompson5910 Would the answer for #2 be about 80.36 meters?

Answer 43

I'm getting h = 80.40816327

Answer 44

double check your work

Answer 45

PE for c = m x g x h
PE for c = 400 kg x 9.8 m/s^2 x 60m
PE for c = 235,200 J

KE for c = 1/2mv^2
KE for c = 1/2(400 kg)(20 m / s)^2
KE for c = 80,000J

KE for b = 0 J

80,000J + 235,200J = 315,000J
PE for b = 315,000J

Answer 46

am i correct so far?

Answer 47

the error made is here

`80,000J + 235,200J = 315,000J`

Answer 48

you should get 80,000J + 235,200J = 315,200J

Answer 49

oh yes

Answer 50

got that :)

Answer 51

then solve 315,200 = 400*9.8*h for h to get h = 80.40816327

Answer 52

for problem 3, how would u find how fast it goes? is it like the velocity?

Answer 53

yeah and you use the KE for that

Answer 54

okay

Answer 55

@jim_thompson5910 how would i find the potential energy for point D?

Answer 56

you did that back in #1

Answer 57

PE = m*g*h

point D is 75 m off the ground

Answer 58

oh no this is another graph. sorry i will attach the picture to u .

Answer 59

So point D is on the ground?

Answer 60

yeah

Answer 61

oh so there isn't any potential energy because distance = 0 meters, right?

Answer 62

but that doesn't make much sense as well

Answer 63

why not? if it's on the ground, then it has no potential to fall further, so it doesn't have any potential energy

Answer 64

so potential energy is equal to 0J?

Answer 65

when it's on the ground, it's all kinetic energy

Answer 66

yeah

Answer 67

oh yes potential energy decreases and kinetic energy increases.

Answer 68

correct

Answer 69

okay thank you so much :)

Answer 70

you're welcome