Speed of a Roller Coaster, Conservation of Energy

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SUMMARY

The discussion focuses on the physics of a roller coaster operating on a frictionless track, specifically analyzing the conservation of energy principles. Given the initial speed of 5.0 m/s at Point A and the heights of Points A (5.0 meters) and C (8.0 meters), the participants derive the necessary equations to determine the speed at Point B and whether the coaster can reach Point C. The key equations utilized include kinetic energy (KE = 0.5*m*v^2) and potential energy (PE = m*g*y), leading to the conclusion that the coaster requires a minimum speed at Point A to reach Point C.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) equations
  • Familiarity with the concepts of conservation of energy in physics
  • Basic knowledge of gravitational acceleration (g = 9.81 m/s²)
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the derivation of energy conservation equations in roller coaster dynamics
  • Learn how to apply the principle of conservation of energy to different mechanical systems
  • Explore the effects of friction on roller coaster speed and energy loss
  • Investigate the role of initial velocity in reaching higher elevations in roller coasters
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in the principles of energy conservation in roller coaster design and dynamics.

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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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