Explaining Hookes Law to a GCSE Physics Student

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SUMMARY

This discussion focuses on Hooke's Law, specifically the equation F = -kx, where F represents the restoring force exerted by a spring, k is the spring constant, and x is the displacement from the equilibrium position. The conversation highlights the relationship between mass and extension, emphasizing that the extension is directly proportional to the applied force until plastic deformation occurs. Additionally, the potential elastic energy (Ee) of a spring is introduced with the formula Ee = 1/2*k*x^2, providing a deeper understanding of the energy dynamics involved.

PREREQUISITES
  • Understanding of basic physics concepts such as force and energy
  • Familiarity with the terms equilibrium position and displacement
  • Knowledge of mathematical equations and their application in physics
  • Basic grasp of potential energy concepts
NEXT STEPS
  • Study the concept of spring constants and their significance in Hooke's Law
  • Learn about potential energy in elastic materials and its calculations
  • Explore the implications of plastic deformation in materials
  • Investigate real-world applications of Hooke's Law in engineering and design
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GCSE Physics students, educators teaching physics concepts, and anyone interested in understanding the principles of elasticity and spring mechanics.

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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