calculusxy
  • calculusxy
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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calculusxy
  • calculusxy
calculusxy
  • calculusxy
@jim_thompson5910
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
  • 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
  • calculusxy
so like: 392,000 = 1/2mv^2
jim_thompson5910
  • jim_thompson5910
yes
calculusxy
  • calculusxy
and then: 392,000 = 1/2(1000)v^2
calculusxy
  • calculusxy
392,000 = 500v^2 784 = v^2 v= 28 m/s
jim_thompson5910
  • jim_thompson5910
very good
calculusxy
  • calculusxy
so acceleration is \[\frac{ \Delta V }{ time }\]
jim_thompson5910
  • jim_thompson5910
over change in time
calculusxy
  • calculusxy
i have to find the two velocities?
calculusxy
  • calculusxy
how would i find the time?
jim_thompson5910
  • jim_thompson5910
the first velocity is 0 since it starts at rest
jim_thompson5910
  • jim_thompson5910
I'm not sure about the time part
calculusxy
  • calculusxy
can i use: distance = 1/2gt^2
jim_thompson5910
  • jim_thompson5910
maybe there's another formula tying together distance, velocity and acceleration
calculusxy
  • calculusxy
distance for point d = 1/2gt^2 75 m = 1/2(9.8m/s^2)t^2
jim_thompson5910
  • jim_thompson5910
I think that's for free fall only?
jim_thompson5910
  • jim_thompson5910
idk if that applies along the curve
calculusxy
  • calculusxy
do u know a formula that i can use to find the time?
jim_thompson5910
  • jim_thompson5910
no I don't sadly
jim_thompson5910
  • jim_thompson5910
do they give any other info?
calculusxy
  • calculusxy
no...
calculusxy
  • calculusxy
but can we do the second question?
jim_thompson5910
  • jim_thompson5910
I'm not sure. They want the acceleration at point F right?
calculusxy
  • calculusxy
yes
jim_thompson5910
  • jim_thompson5910
I'd get a second opinion from a physics expert because I'm not sure about the second part.
calculusxy
  • 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
  • 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
  • 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
  • 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
  • jim_thompson5910
are you referring to problem #1 or #2 ?
calculusxy
  • calculusxy
#1
jim_thompson5910
  • jim_thompson5910
yeah I agree with the work you wrote on the paper (in the attachment)
calculusxy
  • calculusxy
movement = kinetic energy (which doesn't necessarily have to mean acceleration)
jim_thompson5910
  • jim_thompson5910
then you did a bit more work to find that v = 28 m/s
calculusxy
  • calculusxy
yeah i just realized that :)
calculusxy
  • calculusxy
we can work on problem #2 now
jim_thompson5910
  • jim_thompson5910
I wrote out #2 above
calculusxy
  • 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
  • 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
  • calculusxy
@jim_thompson5910 Would the answer for #2 be about 80.36 meters?
jim_thompson5910
  • jim_thompson5910
I'm getting h = 80.40816327
jim_thompson5910
  • jim_thompson5910
double check your work
calculusxy
  • 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
  • calculusxy
am i correct so far?
jim_thompson5910
  • jim_thompson5910
the error made is here `80,000J + 235,200J = 315,000J`
jim_thompson5910
  • jim_thompson5910
you should get 80,000J + 235,200J = 315,200J
calculusxy
  • calculusxy
oh yes
calculusxy
  • calculusxy
got that :)
jim_thompson5910
  • jim_thompson5910
then solve 315,200 = 400*9.8*h for h to get h = 80.40816327
calculusxy
  • calculusxy
for problem 3, how would u find how fast it goes? is it like the velocity?
jim_thompson5910
  • jim_thompson5910
yeah and you use the KE for that
calculusxy
  • calculusxy
okay
calculusxy
  • calculusxy
@jim_thompson5910 how would i find the potential energy for point D?
jim_thompson5910
  • jim_thompson5910
you did that back in #1
jim_thompson5910
  • jim_thompson5910
PE = m*g*h point D is 75 m off the ground
calculusxy
  • calculusxy
oh no this is another graph. sorry i will attach the picture to u .
calculusxy
  • calculusxy
jim_thompson5910
  • jim_thompson5910
So point D is on the ground?
calculusxy
  • calculusxy
yeah
calculusxy
  • calculusxy
oh so there isn't any potential energy because distance = 0 meters, right?
calculusxy
  • calculusxy
but that doesn't make much sense as well
jim_thompson5910
  • 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
  • calculusxy
so potential energy is equal to 0J?
jim_thompson5910
  • jim_thompson5910
when it's on the ground, it's all kinetic energy
jim_thompson5910
  • jim_thompson5910
yeah
calculusxy
  • calculusxy
oh yes potential energy decreases and kinetic energy increases.
jim_thompson5910
  • jim_thompson5910
correct
calculusxy
  • calculusxy
okay thank you so much :)
jim_thompson5910
  • jim_thompson5910
you're welcome

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