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Conservation of energy of a roller coaster car problem

  • Thread starter bmandrade
  • Start date
  • #1
63
0

Homework Statement



A roller coaster car starts from rest and rolls down a frictionless track, reaching Vf ( final speed) at the bottom.

if you want the car to go two times as fast at the bottom by what fator must you increase the height of the track

Homework Equations


KE=1/2mv^2
PE=mgh

The Attempt at a Solution


So according to the conservation of energy
KEi+PEi = KEf+PEf

in this case the KEi = 0 and PEf = 0
so
PEi = KEf
mgh= 1/2mv^2
mass cancels

gh=1/2v^2
2gh=V^2
so v= square root of 2gh

after this i have though to set the equations like this

vf= sqaure root of 2gh
2vf= square root of 2gh

that what i have so far but i dont know how to continue on
 

Answers and Replies

  • #2
367
0
You're almost there: you have

vf=square root of 2gh.

You want a new height H, such that

2vf =square root of 2gH,

so what should H be?
 
  • #3
63
0
ok so this is what i did
i set both equation to vf

so
vf=square root of 2gh
vf= (square root of 2gH)/2

set them equal to each other and divided by 2

2 (square root of 2gh) = suare root of 2gH

(2*(square root of 2gh))^2 = gH

divided by g

((2 *(square root of 2gh))^2)/ g = H

simplified

4*2gh/g = H

g's cancel

8h=H

that means that to increase vf by two we must increse the height by a factor of 8 right???
 
  • #4
367
0
No...but almost: you made a simple mistake. Try again. Can't you see the answer directly by looking at the equations?
 
  • #5
63
0
Ok then it has to increse by a factor of 4!
 
  • #6
367
0
There you go!
 

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