The roller-coaster car shown in Fig. 6-41 (h1 = 30 m, h2 = 8 m, h3 = 28 m), is dragged up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4.
I can't copy and paste the drawing but here is a description of where the points are. Of course h1 would be at the top then the path of the roller coaster declines to point 2 (h2=8 m) and then the coaster assumes to point three which is back up to h3=28 meters.
Here is what I did to calculate the first speed.
v^2=square root of 2xgravityxheight. So......v^2=square root of 2x9.81x30 which = 24.26.
I am having trouble calculating speed 2 and 3 for point 2 and 3. Any help?
2007-02-17 10:00:54 · 5 answers · asked by Jessie L 2 in Science & Mathematics ➔ Physics
No time is provided for this problem and I tried subtracting 8 from thirty using the same formula as suggested but that also yields an incorrect answer.
2007-02-17 11:30:26 · update #1
Think of it like this. What is the potential energy of the roller coaster at it's highest point. The what is the potential energy at the second point. Obviously it will have a higher potential energy when it is at the highest point and will have lost it by the time it gets to the second point. The difference in the two potential energy should be the kinetic energy the roller coaster.
So for h1 what is the potential energy?
P.E.1 = mgh1 = m * 9.8 * 30
What is the potential energy for h2?
P.E.2 = mgh = m * 9.8 * 8
What is the difference in potential energy between the two points?
delta PE12 = PE1 - PE2 = m*9.8*30 - m*9.8*8
We know kinetic energy is 1/2* m *v^2
So set the values to equal each other
detal PE12 = KE12
m*9.8*(30-8) = 1/2*m*v^2
v2 = 20.8 m/sec
Think of this problem like this. At any point on the ride, the roller coaster has potential enery and kinetic energy. The total sum of the two energies do not change along the ride. When the coaster is starting out it has no kinetic energy but lots of potential energy. A the lowest point the the ride, it has no potential energy but lots of kinetic energy, but between the two points the total energy remains the same. So to express this mathematically:
PE1 + KE1 = PE2 + KE2
Can you handle the rest? Message with questions.....
2007-02-17 13:01:12 · answer #1 · answered by nicewknd 5 · 1⤊ 0⤋
Your question is fairly ambiguous, yet i am going to attempt a diverse attitude from the others. Assuming you're in physics, use power. you'll want to attraction to close some assistance about both factors... specifically the speed of the cart at each and each and every aspect and the top large difference between both factors. for sure i am going to ignore about fritction. all of us understand that power is conserved. So the completed power on the initial aspect ought to equivalent the completed power on the most suitable aspect. Ei = Ef the in trouble-free terms energies in contact are Kinetic and ability, so... (KE + PE)i = (KE + PE)f KE = ½m v² PE = mgh considering PE has a nil aspect it is artbitrarily chosen, we may be able to p.c.. PEi = 0, so.. KEi = KEf + PEf ½m vi² = ½m vf² + m g h The m's cancel, and fixing for vf vf = sqrt(vi² - 2gh) be particular to contain the most ideal signal in h... if the coaster is going downhill, h is damaging. If the coaster is going uphill, h is beneficial. There you bypass! desire this helps. If that is confusing, e-mail me or IM me.
2016-12-04 07:36:18 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Roller Coaster Calculations
2017-02-27 16:01:24 · answer #3 · answered by ? 4 · 0⤊ 0⤋
the "h" in the formula mgh, refers to the change in height and therfore at h2 it should be, m(9.81)(30 - 8) = mv^2 and likewise for the other points h3 and h4
2007-02-17 10:48:29 · answer #4 · answered by human 2 · 2⤊ 0⤋
from the formula you provided, v²=√2gh (g=9.8m/s²), just plug in h2 and h3.
but the formula for velocity is v=gt when initial velocity is 0, so there has to be a time provided.
2007-02-17 10:43:39 · answer #5 · answered by Winnie 3 · 0⤊ 0⤋