Speed of a Roller Coaster, Conservation of Energy

AI Thread Summary
The discussion centers on calculating the speed of a roller coaster at different points using the principles of conservation of energy. The initial speed at Point A is given as 5.0 m/s, and the heights of Points A and C are 5.0 meters and 8.0 meters, respectively. Participants clarify that the kinetic energy (KE) at Point A plus the potential energy (PE) must equal the KE at Point B, leading to the equation m*g*h + 0.5*m*u^2 = 0.5*m*v^2. The conversation also addresses whether the roller coaster can reach Point C and what initial speed is necessary for that. The thread concludes with a request for further assistance due to confusion over the calculations.
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Homework Statement



A roller coaster travels on a frictionless track as shown in the figure.
(a) If the speed of the roller coaster at Point A is 5.0 m/s, what is its speed at Point B?
(b) Will it reach Point C?
(c) What speed at Point A is required for the roller coaster to reach Point C?

Please view the attached figure. In case the image is too blury, the height of Point A = 5.0 meters, and the height of Point C = 8.0 meters.

Homework Equations



KE = 0.5*m*v^2
PE = m*g*y, where y = vertical position

The Attempt at a Solution



KE = PE
0.5*m*v^2 = m*g*y
mass cancels
0.5*v^2 = g*y
v = √(2*g*y)

What confuses me is what to do with the starting speed at Point A, 5.0 m/s. If the velocity at the top was zero, then the problem would be relatively straightforward. Where do I go from here? Thanks in advance!
 

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A) PE at point a and KE = KE at point b
So m*g*h + 0.5*m*u^2 = 0.5*m*v^2
I think you can do the rest
B) Nope
C) KE lost = PE gained
I can't see the numbers on the image attached!
 
alewisGB said:
A) PE at point a and KE = KE at point b
So m*g*h + 0.5*m*u^2 = 0.5*m*v^2
I think you can do the rest
B) Nope
C) KE lost = PE gained
I can't see the numbers on the image attached!

Excellent, thank you!
 
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