Hookes Law and Spring Constant: Explaining the Integral

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SUMMARY

This discussion centers on Hooke's Law and the spring constant "k," specifically addressing the integral U = 1/2 kx^2. The user initially misunderstands the derivation of the area under the force versus displacement graph, which is a triangle representing the work done on the spring. The correct interpretation shows that the area, calculated as (1/2) * base * height, leads to the integral of the force function F = kx, resulting in the expression (kx^2)/2 as the primitive function.

PREREQUISITES
  • Understanding of Hooke's Law and its formula F = kx
  • Basic knowledge of calculus, specifically integration
  • Familiarity with the concept of area under a curve
  • Ability to interpret graphical representations of mathematical functions
NEXT STEPS
  • Study the derivation of the work done on a spring using calculus
  • Learn about the graphical interpretation of integrals in physics
  • Explore advanced applications of Hooke's Law in mechanical systems
  • Investigate the relationship between force, displacement, and energy in elastic materials
USEFUL FOR

Students of physics, engineers, and anyone interested in understanding the mathematical foundations of spring mechanics and energy calculations.

Alec
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Hello, I'm having some problems with the spring constant "k" together with Hookes law. U=1\2kx^2

Could someone please explain how you get that integral?
If you insert it in a diagram and calculate the area as a triangle you would get 1/2kx. Where does the ^2 come from?
 
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No, you don't!
Draw the triangle again.
How long is its base, and how long is its height?
 
Alright, my bad :rolleyes: I missed that it's (kx * x)/2
Now, another question.
The function for F = kx.
The primitiv function for kx = (kx^2)/2 right? Because then you could calculate it as an integral?
 

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