Difference between hooke's law and the work done in a spring?

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SUMMARY

The discussion clarifies the distinction between Hooke's Law and the work done on a spring. Hooke's Law states that the spring force is given by F = -kx, where k is the spring constant and x is the displacement. The work done in stretching a spring is calculated using the formula W = 1/2 kx^2, which accounts for the variable force exerted by the spring as it stretches. The factor of 1/2 arises because the force is not constant during the stretching process, necessitating the use of calculus to integrate the force over the distance.

PREREQUISITES
  • Understanding of Hooke's Law and its mathematical representation (F = -kx)
  • Basic knowledge of work and energy concepts in physics
  • Familiarity with calculus, particularly integration
  • Ability to differentiate between constant and variable forces
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  • Study the principles of calculus, focusing on integration techniques
  • Explore the concept of variable forces in physics
  • Learn about energy conservation in mechanical systems
  • Investigate applications of Hooke's Law in real-world scenarios
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dnt
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hooke's law says the spring force = -kx

but the work done in stretching a spring = 1/2 kx^2

isnt work = Fd?

so using W=Fd and F=-kx (hooke's law) shouldn't the work come out to:

W = -kx^2? (arent d and x the same? both are distance stretched)

where does the 1/2 come from in the equation 1/2kx^2?

can someone help clarify the difference and when to use each equation. thanks.
 
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dnt said:
hooke's law says the spring force = -kx

but the work done in stretching a spring = 1/2 kx^2

isnt work = Fd?

so using W=Fd and F=-kx (hooke's law) shouldn't the work come out to:

W = -kx^2? (arent d and x the same? both are distance stretched)

where does the 1/2 come from in the equation 1/2kx^2?

can someone help clarify the difference and when to use each equation. thanks.
W = Fd is only correct when the force is constant and in the direction of a displacement of distance d. The force is not constant for a spring. Have you taken calculus?
 
dnt said:
isnt work = Fd?
That is correct, provided the force is constant. However, generally work can be expressed thus;

W=\int^{x_{1}}_{x_{0}}\; F \cdot dx

Edit: Too slow again :frown:
 
Hootenanny said:
That is correct, provided the force is constant. However, generally work can be expressed thus;

W=\int^{x_{1}}_{x_{0}}\; F \cdot dx

Edit: Too slow again :frown:

...which, after plugging in F = - k x, answers your question. :smile:
 

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