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.

- calculusxy

- katieb

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- calculusxy

##### 1 Attachment

- calculusxy

- calculusxy

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

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## More answers

- jim_thompson5910

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

- calculusxy

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

- jim_thompson5910

yes

- calculusxy

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

- calculusxy

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

- jim_thompson5910

very good

- calculusxy

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

- jim_thompson5910

over change in time

- calculusxy

i have to find the two velocities?

- calculusxy

how would i find the time?

- jim_thompson5910

the first velocity is 0 since it starts at rest

- jim_thompson5910

I'm not sure about the time part

- calculusxy

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

- jim_thompson5910

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

- calculusxy

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

- jim_thompson5910

I think that's for free fall only?

- jim_thompson5910

idk if that applies along the curve

- calculusxy

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

- jim_thompson5910

no I don't sadly

- jim_thompson5910

do they give any other info?

- calculusxy

no...

- calculusxy

but can we do the second question?

- jim_thompson5910

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

- calculusxy

yes

- jim_thompson5910

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

- calculusxy

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?

- jim_thompson5910

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

- phi

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)

- calculusxy

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?

- jim_thompson5910

are you referring to problem #1 or #2 ?

- calculusxy

#1

- jim_thompson5910

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

- calculusxy

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

- jim_thompson5910

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

- calculusxy

yeah i just realized that :)

- calculusxy

we can work on problem #2 now

- jim_thompson5910

I wrote out #2 above

- calculusxy

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?

- jim_thompson5910

`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

- calculusxy

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

- jim_thompson5910

I'm getting h = 80.40816327

- jim_thompson5910

double check your work

- calculusxy

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

- calculusxy

am i correct so far?

- jim_thompson5910

the error made is here
`80,000J + 235,200J = 315,000J`

- jim_thompson5910

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

- calculusxy

oh yes

- calculusxy

got that :)

- jim_thompson5910

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

- calculusxy

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

- jim_thompson5910

yeah and you use the KE for that

- calculusxy

okay

- calculusxy

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

- jim_thompson5910

you did that back in #1

- jim_thompson5910

PE = m*g*h
point D is 75 m off the ground

- calculusxy

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

- calculusxy

##### 1 Attachment

- jim_thompson5910

So point D is on the ground?

- calculusxy

yeah

- calculusxy

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

- calculusxy

but that doesn't make much sense as well

- jim_thompson5910

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

- calculusxy

so potential energy is equal to 0J?

- jim_thompson5910

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

- jim_thompson5910

yeah

- calculusxy

oh yes potential energy decreases and kinetic energy increases.

- jim_thompson5910

correct

- calculusxy

okay thank you so much :)

- jim_thompson5910

you're welcome

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