What is the final velocity of a particle in a force field around point O?

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A particle with mass m experiences a force F around point O, defined as F = cos(d/5), and begins at rest from a negligible distance away. To find the final velocity, one must calculate the work done on the particle while moving it to distance d from O. The term "negligible distance" allows for zero to be used as the lower limit in integration, indicating the particle starts effectively from 0. This setup implies that the particle moves from 0 to d, which is the intent of the problem. Understanding these parameters is essential for determining the particle's final velocity.
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A force field is maintained around point O, a particle with mass m is experiencing a force F in the force field. F as a function of the particle's distance from O is: F = cos(d/5) How does one go about looking for the final velocity of the particle if it began at rest at a negligible distance away from point O?
 
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Calculate the work done on the particle by the force in moving it to distance d away from O. Then find the velocity at which the mass has that kinetic energy.
 
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andrewkirk said:
Calculate the work done on the particle by the force in moving it to distance d away from O. Then find the velocity at which the mass has that kinetic energy.
Absolutely. That is the way to go. The only question I have is, what exactly does the statement of the problem mean by the word, "negligible distance".?
 
In practice it means you can use zero as the lower limit for your integration. I expect the reason they said 'negligible distance' rather than 'start at O' is that usually when forces are symmetrically arranged around a point it's because there's a particle at that point, and two particles can't occupy the same point in classical phycics.
 
andrewkirk said:
In practice it means you can use zero as the lower limit for your integration. I expect the reason they said 'negligible distance' rather than 'start at O' is that usually when forces are symmetrically arranged around a point it's because there's a particle at that point, and two particles can't occupy the same point in classical phycics.
Ah! So the particle is moving effectively from 0 to d?
 
Yes, I think that is the intent of the question.
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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