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

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Homework Help Overview

The discussion revolves around the concepts of Hooke's Law and the work done on a spring. Participants are exploring the relationship between the force exerted by a spring and the work required to stretch or compress it, particularly focusing on the equations F = -kx and W = 1/2 kx^2.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the application of the work formula W = Fd in the context of a spring, noting that the force is not constant during stretching. They are also discussing the origin of the 1/2 factor in the work done equation.

Discussion Status

The discussion is active, with participants providing insights into the nature of variable forces and the integral approach to calculating work. Some have pointed out the need for calculus to fully understand the work done on a spring.

Contextual Notes

There is an indication that some participants may not have a background in calculus, which is relevant to the discussion of variable forces and the integration of force over distance.

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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