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

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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
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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?

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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
 
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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
 
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