Conservation of Energy roller-coaster car

AI Thread Summary
The discussion focuses on calculating the minimum speed v0 required for a roller-coaster car to ascend a hill after descending a valley, using conservation of energy principles. The participants agree that since there is no friction or air resistance, the total mechanical energy is conserved, leading to the equation Ui + Ki = Uf + Kf. They explore the heights involved, with H1 at 4.1 m and H2 at 4.6 m, and consider whether to use one or two equations to solve for v0. A participant attempts to derive the speed but questions their calculation, indicating a need for clarification on the energy conservation approach. The discussion highlights the importance of correctly applying energy conservation in roller-coaster physics.
maniacp08
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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)
 
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maniacp08 said:
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
 
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?
 
Can someone help me on this one?
 

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