How Does the Cyclone Roller Coaster at Six Flags Use Physics Principles?

  • Thread starter Thread starter rottentreats64
  • Start date Start date
  • Tags Tags
    Energy Power Work
AI Thread Summary
The Cyclone roller coaster at Six Flags in Massachusetts features a 34.1m ascent and a 21.9m drop, with a train mass of 4727kg. The work required to elevate the empty train to the first hill is calculated to be approximately -1,611,907 joules. Power needed to reach the top in 30 seconds is determined to be about 6,125.25 watts. To find the potential energy (PE) converted into kinetic energy (KE) during the drop, the height difference must be considered, as PE is dependent on the height relative to another point. Understanding these physics principles is crucial for analyzing roller coaster dynamics.
rottentreats64
Messages
19
Reaction score
0
4. At Six Flags in Massachusetts, a ride called the Cyclone is a giant roller coaster that ascends a 34.1m hill and then drops 21.9m before ascending the next hill. The train of cars has a mass of 4727kg.

a. how much work is required to get an empty train of cars from the ground to the top of the first hill?

b. what power must be generated to bring the train to the top of the first hill in 30.0s?

c. how much PE is converted into KE from the top of the first hill to the bottom of the 21.9m drop?

-----------------------------------------------------------------------------
4. For this problem I used w=f*d but I needed to find out the force to find the work. So I did f=m*aà f=(4727)*-10(gravity) -47270. Then I just plugged in the force for the work equation and found out that the work required is -1611907 j .For part b I just used the power equation (work/time) therefore (5388.78*34.1)/(30) and you get the answer 6125.2466. For part c i don't know what to do
 
Physics news on Phys.org
rottentreats64 said:
4. At Six Flags in Massachusetts, a ride called the Cyclone is a giant roller coaster that ascends a 34.1m hill and then drops 21.9m before ascending the next hill. The train of cars has a mass of 4727kg.

a. how much work is required to get an empty train of cars from the ground to the top of the first hill?

b. what power must be generated to bring the train to the top of the first hill in 30.0s?

c. how much PE is converted into KE from the top of the first hill to the bottom of the 21.9m drop?

-----------------------------------------------------------------------------
4. For this problem I used w=f*d but I needed to find out the force to find the work. So I did f=m*aà f=(4727)*-10(gravity) -47270. Then I just plugged in the force for the work equation and found out that the work required is -1611907 j .For part b I just used the power equation (work/time) therefore (5388.78*34.1)/(30) and you get the answer 6125.2466. For part c i don't know what to do

If you are going to round off g to 10 you are going to have to round it off to 1.6e6 J.

For part c, what is the height difference between the two points? What is the PE of a mass at height h relative to another point?

AM
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top