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calculusxy
 one year ago
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
 one year ago
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.

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calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1So would this be related to the Law of Conservation of Energy: (PE + KE) = (PE + KE)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I think once you get 392,000 J for the KE at point F, you use that to find the velocity v

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1so like: 392,000 = 1/2mv^2

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1and then: 392,000 = 1/2(1000)v^2

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1392,000 = 500v^2 784 = v^2 v= 28 m/s

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1so acceleration is \[\frac{ \Delta V }{ time }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1over change in time

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1i have to find the two velocities?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1how would i find the time?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the first velocity is 0 since it starts at rest

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm not sure about the time part

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1can i use: distance = 1/2gt^2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1maybe there's another formula tying together distance, velocity and acceleration

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1distance for point d = 1/2gt^2 75 m = 1/2(9.8m/s^2)t^2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I think that's for free fall only?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1idk if that applies along the curve

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1do u know a formula that i can use to find the time?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1no I don't sadly

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1do they give any other info?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1but can we do the second question?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm not sure. They want the acceleration at point F right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'd get a second opinion from a physics expert because I'm not sure about the second part.

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.1PE 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
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1are you referring to problem #1 or #2 ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah I agree with the work you wrote on the paper (in the attachment)

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1movement = kinetic energy (which doesn't necessarily have to mean acceleration)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then you did a bit more work to find that v = 28 m/s

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1yeah i just realized that :)

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1we can work on problem #2 now

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I wrote out #2 above

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1oh 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
 one year ago
Best ResponseYou've already chosen the best response.1`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
 one year ago
Best ResponseYou've already chosen the best response.1@jim_thompson5910 Would the answer for #2 be about 80.36 meters?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm getting h = 80.40816327

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1double check your work

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1PE 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
 one year ago
Best ResponseYou've already chosen the best response.1am i correct so far?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the error made is here `80,000J + 235,200J = 315,000J`

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you should get 80,000J + 235,200J = 315,200J

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then solve 315,200 = 400*9.8*h for h to get h = 80.40816327

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1for problem 3, how would u find how fast it goes? is it like the velocity?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah and you use the KE for that

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1@jim_thompson5910 how would i find the potential energy for point D?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you did that back in #1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1PE = m*g*h point D is 75 m off the ground

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1oh no this is another graph. sorry i will attach the picture to u .

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1So point D is on the ground?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1oh so there isn't any potential energy because distance = 0 meters, right?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1but that doesn't make much sense as well

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1why not? if it's on the ground, then it has no potential to fall further, so it doesn't have any potential energy

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1so potential energy is equal to 0J?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1when it's on the ground, it's all kinetic energy

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1oh yes potential energy decreases and kinetic energy increases.

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.1okay thank you so much :)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you're welcome
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