Qns on potential energy and forces

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Homework Help Overview

The discussion revolves around the concept of potential energy and the forces acting on a body in a physics context. The original poster presents a question regarding the force acting on a body at a specific point based on a given potential energy function.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the force from the potential energy equation and expresses confusion regarding the correct answer. Some participants discuss the relationship between work, force, and potential energy, while others clarify the implications of variable forces.

Discussion Status

Participants are exploring different interpretations of the relationship between potential energy and force. Some have provided insights into the mathematical derivation of force from potential energy, while others have noted the importance of the reference point in determining potential energy. There is no explicit consensus, but the discussion is progressing with valuable contributions.

Contextual Notes

The original poster's understanding of the force derived from potential energy appears to be challenged, indicating a need for further clarification on the concepts involved. The discussion includes considerations of how potential energy is defined relative to a reference point.

coffeebeans
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Hello, i hope i posted this in the correct section..

I have a question:
the potential energy of a body when it is at point P a distance x from a reference point O is given by V = kx^2, where k is a constant. what is the force acting on the body when it is at P?

the correct ans is 2kx in the direction of PO

the ans i got is kx in the direction OP. as work = force x distance, so force required to bring the body to point p is kx^2 / x .

could some please explain why is the correct ans so? thnk u so much!
 
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Work is "force times distance" as long as the force is constant- if the force is a variable then the work is given by [itex]\int f(x)dx[/itex]. Notice that this involves a "constant of integration". It is standard to choose that constant so that potential energy is 0 at some specific point; potential energy is always relative to some given point.

Going the other way, if work is a constant times distance, W= Cx then force is that constant: F= Cx/x= C. But if a more general function then [itex]F= \frac{dW}{dx}[/itex].
 
The potential energy is positive. So there is an attraction toward the origin.

The force = - (dW/dx) = -2kx.
So the force is directed from the point P to the point O, the origin.
bye
 
Thnx a lot!
 

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