Find the potential energy function

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SUMMARY

The discussion focuses on deriving the potential energy function \( U(x) \) for a particle constrained to move along the x-axis under the influence of a force \( \vec F(x) = \frac{-k}{x^3} \vec i \). The correct potential energy function is established as \( U(x) = \frac{k}{8} - \frac{k}{2x^2} \), with the condition that \( U(2.0 \, \text{m}) = 0 \). The misunderstanding arose from the incorrect application of the integral in the potential energy equation, where the negative integral of the force must be used.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly force and potential energy.
  • Familiarity with calculus, specifically integration techniques.
  • Knowledge of the relationship between work and potential energy.
  • Basic understanding of SI units and constants in physics.
NEXT STEPS
  • Study the derivation of potential energy functions from force equations in classical mechanics.
  • Learn about the application of definite integrals in physics problems.
  • Explore the implications of potential energy in conservative force fields.
  • Investigate the role of boundary conditions in defining potential energy functions.
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Students of physics, particularly those studying classical mechanics, as well as educators and tutors looking to clarify concepts related to force and potential energy functions.

Calpalned
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Homework Statement


A particle is constrained to move in one dimension along the x-axis and is acted upon by a force given by ##\vec F (x) = \frac{-k}{x^3} \vec i ## where ##k ## is a constant with units appropriate to the SI system. Find the potential energy function ##U(x)##, if U is arbitrarily defined to be zero at x = 2.0 m, so that ##U(2.0 m) = 0 ##

Homework Equations


## \Delta U = -W = \int \vec F * dl ##

The Attempt at a Solution


The textbook says that answer is ##U(x) = \frac {k}{8} - \frac{k}{2x^2} ## but I got ##U(x) = \frac{k}{2x^2} - \frac {k}{8} ## Am I still correct?
 
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Well, you're not far off, except the equation for Potential in the relevant equations is the negative integral of the force. Not just the integral. There's your issue.
 
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