Potential energy particle problem

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SUMMARY

The discussion focuses on calculating the potential energy associated with a conservative force defined as F = (4.0x - 13)i N acting on a particle along the x-axis. The potential energy U is derived from the force using the equation U = -∫F dx, resulting in U = 2x² - 13x + c. At x = 0, U is assigned a value of 26 J, leading to the determination of the maximum potential energy and the specific x-values where potential energy equals zero. The critical points are identified at x = 13/4, indicating a minimum rather than a maximum.

PREREQUISITES
  • Understanding of conservative forces and potential energy concepts
  • Knowledge of calculus, specifically integration and differentiation
  • Familiarity with quadratic functions and their properties
  • Ability to analyze turning points in mathematical functions
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  • Study the principles of conservative forces and their relation to potential energy
  • Learn how to perform integration of force functions to derive potential energy equations
  • Explore the characteristics of quadratic functions and their maxima/minima
  • Investigate the physical significance of turning points in the context of energy conservation
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Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for examples of potential energy calculations in conservative force systems.

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Homework Statement



A single conservative force F = (4.0x - 13)i N, where x is in meters, acts on a particle moving along an x axis. The potential energy U associated with this force is assigned a value of 26 J at x = 0. (a) What is the maximum positive potential energy? At what (b) negative value and (c) positive value of x is the potential energy equal to zero?

Homework Equations



Force = dU/dx where U is the potential energy.

The Attempt at a Solution



Okay, so we find the anti-derivative of the given force then we have: U = x^2/2 - 13x. Then what? At x = 0 m, U is 26 J. At the maximum U, we know the kinetic energy is 0 (v = 0).
 
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note:
force is zero when x=13/4, everywhere else it points away from there.

the indefinite integral of 4x-13 is actually

2x2-13x+c

Which is a quadratic (c is the constant of integration).

The turning point, is where the derivative is equal to zero :) though, in this case, it looks like a minima rather than the asked-for maxima (check the equation does not have a minus sign in front).
 

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