How Do You Calculate the Kinetic Energy of a Roller Coaster?

[anna sung]

Homework Statement

the rollar coaster is at a height of 15m and going up hill of 47m. its mass is 4.0 x 10^4kg.
find the kinemetic energy of it.

Homework Equations

Ek + Eg = Ek2 + Eg2
PE= mgh
KE= 1/2mv^2

The Attempt at a Solution

i find the potential energy of it which is
PE= 15(9.80)(4.0 X 10^4)
= 5.9 X 10^6 J
PE2= 47(9.80)(4.0 X 10^4)
= 1.8 x 10^7 J

and i said KE + PE = KE2 +PE2
0J + 5.9 X 10^6J = KE2 + 1.8 X 10^7J

the answer is 2.4 x 10^6 J
please tell me what i am doing wrong i am so confused.

 

[PhanthomJay]

Anna ----you have not described the problem completely. There is missing information. For example, why did you choose the iinitial kinetic energy at elevation 15 to be 0??

 

[anna sung]

uh. i don't know that's the only information that the text is given us. it gave us the mass,
and the roller coaster starts from 15m going up a hill of 47m.
and asked to calculate the kinetic energy.

 

[PhanthomJay]

Was there a sketch? Was the initial height of the coaster given when it started its first drop from the top when at rest? Is the problem asking for KE at 47 m or KE at 15 m? Is the missing info the data from the other problem you asked about, where the coaster height at rest is 54 m??

 

[anna sung]

oh yes yes. sorry it starts at 54m. and go down to 15m then up again 47m.
yes. there is a sketch. i wanted to asked for only one part of it. cause if i know how to do
point c to b . then i could do the rest.

 

[PhanthomJay]

Well, that problem is asking for the kinetic and potential energy of the coaster at a given point. Once you know the kinetic and potential energy at one point, and the PE at another point, you can use the conservation of energy equation to solve for the KE at that other point. The KE at elevation 15 can't be zero, that means it would have no speed at that point. Instead, start at point A (the top of the coaster ride) where you know the PE and you should know the KE at that point A where the coaster is at rest. Then continue from there.