Integration of an interaction force to find potential energy

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SUMMARY

The discussion centers on calculating the potential energy from the interaction force Fx=(3x²−5x) N acting on a particle along the x-axis. The correct expression for potential energy is derived from integrating the force, resulting in U(x) = -∫(3x²−5x) dx = -x³/3 + 5/2 x² + C, where C is the constant of integration. A critical aspect highlighted is the importance of the negative sign in the relationship between force and potential energy, as expressed by F(x) = -dU/dx, which is essential for accurate calculations.

PREREQUISITES
  • Understanding of basic calculus, specifically integration techniques.
  • Familiarity with the relationship between force and potential energy in physics.
  • Knowledge of Hooke's law and its implications for potential energy.
  • Ability to manipulate algebraic expressions involving constants of integration.
NEXT STEPS
  • Study the derivation of potential energy from force using integration techniques.
  • Explore the implications of Hooke's law on potential energy calculations.
  • Learn about the significance of the negative sign in the force-potential energy relationship.
  • Investigate other forms of potential energy in different physical systems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking to clarify the relationship between force and potential energy.

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Homework Statement


A particle that can move along the x-axis experiences an interaction force Fx=(3x2−5x) N where x is in m. Find an expression for the system's potential energy. Express your answer in terms of the variables x and the constant of integration C, where C is in joules.

Homework Equations


I used simple integration rules to solve this problem.

The Attempt at a Solution


∫(3x2−5x) = x3-5/2 x2+C

I put this answer and the feedback said to "check your signs". I was confident that the above was right, so where is my mistake in the sign?
 
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You did the integral correctly, but you're missing the physics bit of it.

Is the potential energy really just given by the integral of the force? Try it on Hooke's law.

F = -kx ---> U(x) = -1/2 kx^2? Does this seem right?

The punch here is that, for F = F(x)

F(x) = -\frac{dU}{dx}

note the minus sign
 
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