Explaining Hookes Law to a GCSE Physics Student

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Discussion Overview

The discussion revolves around explaining Hooke's Law to a GCSE Physics student, focusing on the mathematical formulation of the law and its implications regarding springs and potential elastic energy. The scope includes conceptual clarification and basic mathematical reasoning.

Discussion Character

  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • A GCSE Physics student expresses confusion about the equation F = -kx, seeking clarification on its components and meaning.
  • Another participant explains that a spring can be stretched or compressed and that this deformation relates to potential elastic energy when the spring is not at equilibrium.
  • The same participant notes that k is a constant specific to the spring, while x represents the distance from the equilibrium position.
  • A formula for potential elastic energy, Ee = 1/2*k*x^2, is introduced as relevant to the discussion.

Areas of Agreement / Disagreement

The discussion appears to be collaborative, with participants attempting to clarify concepts, but there is no explicit consensus on the student's understanding of Hooke's Law or the equations involved.

Contextual Notes

The discussion does not address potential limitations in the student's prior knowledge or assumptions about the concepts being discussed.

cwilloughby21
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I am a GCSE Physics Student I was hoping somebody would be able to help me when I have been studying Hookes law, we have only looked at a basic equation extension = extented length - original length. We have also look at how the increase in mass is directly proportional to the extension until plastic deformation occurs. I was then looking through my revsion notes and the equation F=-KX came up. I looked this up and found that
athematically, Hooke's law states that

F=-kx,

where

x is the displacement of the spring's end from its equilibrium position (a distance, in SI units: meters);
F is the restoring force exerted by the spring on that end (in SI units: N or kg·m/s2); and
k is a constant called the rate or spring constant (in SI units: N/m or kg/s2).

this makes no sense to me
Could somebody please explain this to me
thanks
 
Last edited:
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Hey, I'm in gr 12 so I don't know how much help I am, but I can try. Think about a spring for one second:

A spring can be stretched or compressed. When a spring is stretched, or compressed, it has POTENTIAL ELASIC ENERGY. When the spring is at eqilbrium postion, that means it has not been stretched or compressed therefore it has no potential elastic energy. K is just a constant ( a given number) that the spring has. Finally, x is the distance the springs been stretched or compressed FROM EQILBRIUM POSTION.

If you have anymore questions just ask, I hope this helped.
 
cwilloughby21 said:
this makes no sense to me
Why doesn't that make sense?
 
Oh yeah, Ee (which is potential elastic energry) = 1/2*k*x^2 this is formula you may use.
 
Anythin we could add Doc?
 
thanks hellohi that was really helpful :smile:
 
Np man, I'm glad it helped:)!
 

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