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

[physixnot4me]

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.

 

[ZapperZ]

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.

 

[quicknote]

Hello,

Thank you for sharing the diagram and information regarding the potential energy of the conservative force acting on the particle. From the given information, we can use the concept of conservation of energy to determine the position of the left turning point for the particle.

At t=0, the particle is at rest and has a potential energy of 4J at x=12. As the particle moves under the influence of the conservative force, its kinetic energy will increase and its potential energy will decrease. However, the total energy (kinetic + potential) of the particle will remain constant.

Using this concept, we can set up an equation to find the position of the left turning point:

Initial energy (at t=0) = Final energy (at left turning point)

4J = kinetic energy + potential energy at left turning point

Since the particle is at rest at t=0, its kinetic energy is zero. Therefore, we can rewrite the equation as:

4J = 0 + potential energy at left turning point

Solving for the potential energy at the left turning point, we get:

Potential energy at left turning point = 4J

Since we know that the potential energy at x=12 is also 4J, this means that the left turning point is at x=12. This is the point where the particle stops and changes direction, as the potential energy is at its minimum at this point.

I hope this helps. Let me know if you have any further questions. Thank you.