What Is the Left Turning Point of a Particle in a Conservative Force Field?

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SUMMARY

The left turning point of a particle in a conservative force field can be determined using the principle of conservation of energy. Initially, at position x=12, the particle has potential energy (PE) of 4J and no kinetic energy (KE). As the particle moves, it will convert PE to KE until it reaches a point where all energy is converted back to PE, at which point it will momentarily stop before reversing direction. This specific position on the potential energy graph indicates the left turning point.

PREREQUISITES
  • Understanding of conservative force fields
  • Knowledge of potential energy (PE) and kinetic energy (KE)
  • Familiarity with energy conservation principles
  • Ability to interpret potential energy graphs
NEXT STEPS
  • Study the conservation of energy in conservative systems
  • Learn how to analyze potential energy graphs
  • Explore the concept of turning points in mechanics
  • Investigate the relationship between PE and KE in motion
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Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of particles in conservative force fields.

physixnot4me
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2. at t=0 the particle is at rest x=12. after is release it can move under influence of conservative force whose potential energy is shown in the diagram. what is the position of the left turning point for this particle?

ans: At the moment, I am not sure how to convert his graph to visualize the position of the particle, in order to find the left turning point. All i see is at x=12, the U is 4J. what can i do with this?

attached is the diagram.
 

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physixnot4me said:
2. at t=0 the particle is at rest x=12. after is release it can move under influence of conservative force whose potential energy is shown in the diagram. what is the position of the left turning point for this particle?
ans: At the moment, I am not sure how to convert his graph to visualize the position of the particle, in order to find the left turning point. All i see is at x=12, the U is 4J. what can i do with this?
attached is the diagram.

Use conservation of energy. At x=12, it has all of its energy in the form of PE, and no KE. Upon release, as it is moving through the different positions, it will have a combination of PE and KE (think of a rollercoaster). As long as PE (U in that diagram) is less than the original value, there will be KE and the particle will continue to move. There is only one location in that graph where the particle will regain all of the original PE. At this point, it will no longer have any KE and it will momentarily stop, before reversing its motion. That's the turning point.

I don't think I need to tell you where this point is, do I?

Zz.
 

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