Conservation of Energy roller-coaster car

  • Thread starter maniacp08
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  • #1
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A single roller-coaster car is moving with speed v0 on the first section of track when it descends a 4.1-m-deep valley, then climbs to the top of a hill that is 4.6 m above the first section of track. Assume any effects of friction or of air resistance are negligible.

(a) What is the minimum speed v0 required if the car is to travel beyond the top of the hill?

Since there is no friction or air resistance then all the energy is conserved.
There is no external work so the equation is:
Ui + Ki = Uf + Kf

H1 = 4.1m
H2 = 4.6m
If I take y=0 on the first section of track.

Would I need to do the equation twice since there are two Heights involved?
Like Ui = 0 and Uf be MG(-H1)

2nd equation
Ui = MG(-H1) and Uf be MG(H2)
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,097
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A single roller-coaster car is moving with speed v0 on the first section of track when it descends a 4.1-m-deep valley, then climbs to the top of a hill that is 4.6 m above the first section of track. Assume any effects of friction or of air resistance are negligible.

(a) What is the minimum speed v0 required if the car is to travel beyond the top of the hill?

Since there is no friction or air resistance then all the energy is conserved.
There is no external work so the equation is:
Ui + Ki = Uf + Kf

H1 = 4.1m
H2 = 4.6m
If I take y=0 on the first section of track.

Would I need to do the equation twice since there are two Heights involved?
Like Ui = 0 and Uf be MG(-H1)

2nd equation
Ui = MG(-H1) and Uf be MG(H2)

You can do the 2 equations, but note that it might as well be just the one that calculates the
KE = 1/2mv2 = m*g*(H2 - Ho) = Increase in PE
 
  • #3
115
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So I can do this in one equation?
Ui + Ki = Uf + Kf
0 + 1/2 M * Vi^2 = m*g*(H2 - H1) + 0

1/2 M * Vi^2 = m*g*(H2 - H1)
= Vi^2 = 9.81
Vi = square root of 9.81 = 3.1 m/s?

This answer is wrong tho. Anything I did wrong?
 
  • #4
115
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Can someone help me on this one?
 

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