Finding speed at half the height given initial velocity and mass

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SUMMARY

The discussion centers on calculating the speed of a frictionless roller coaster car with a mass of 800 kg at a height of 22 m (half of the initial height of 44 m) given an initial speed of 18 m/s. The initial kinetic energy (KE) is calculated as 129600 J, and the potential energy (PE) at the top is 3520 J. The total energy remains constant, leading to the conclusion that the speed at point B is 12.7 m/s. The misconception that kinetic energy is simply halved when height is halved is clarified by emphasizing the conservation of total energy.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) concepts
  • Familiarity with the conservation of energy principle
  • Basic knowledge of physics equations related to motion
  • Ability to perform calculations involving mass, height, and speed
NEXT STEPS
  • Study the conservation of mechanical energy in frictionless systems
  • Learn how to calculate potential energy (PE) and kinetic energy (KE) in various scenarios
  • Explore the effects of friction on energy conservation in roller coaster dynamics
  • Investigate real-world applications of energy conservation in engineering
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of energy conservation in mechanical systems, particularly in the context of roller coaster dynamics.

rockchalk1312
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In the figure, a frictionless roller coaster car of mass m = 800 kg tops the first hill with speed v0 = 18 m/s at height h = 44 m. What is the speed of the car at point B?

(point B is one half the height of the top of the first hill)


(1/2)mv2
Wg=mgd


(1/2)(800)(182)=129600
129600/2=64800=(1/2)(800)(v2)
v=12.7m/s

This was wrong. Is the kinetic energy not just cut in half when height is halved? Thank you!
 
Last edited:
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rockchalk1312 said:
In the figure, a frictionless roller coaster car of mass m = 800 kg tops the first hill with speed v0 = 18 m/s at height h = 44 m. What is the speed of the car at point B?

(point B is one half the height of the top of the first hill)


(1/2)mv2
Wg=mgd


(1/2)(800)(182)=129600
129600/2=64800=(1/2)(800)(v2)
v=12.7m/s

This was wrong. Is the kinetic energy not just cut in half when height is halved? Thank you!

It's hard to decipher what you are doing. You know the total energy TE is the sum of the PE and KE, and it is constant at all times because of the frictionless track. Write the explicit sum for each of the points on the track...
 

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