How Do I Calculate the Spring Constant from a Linear Equation?

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Homework Help Overview

The discussion revolves around calculating the spring constant from a linear equation derived from experimental data involving force and extension of a spring-like system. The original poster presents a linear equation obtained from a graph of force versus extension.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the slope of their linear equation to find the spring constant but encounters inconsistencies in their results. Some participants clarify the relationship between force and extension, suggesting that the gradient of the line represents the spring constant.

Discussion Status

The discussion is ongoing, with participants providing clarifications about the relationship between the slope of the graph and the spring constant. There is an acknowledgment of the simplicity of the concept, but no explicit consensus has been reached regarding the original poster's calculations.

Contextual Notes

The original poster's equation includes a y-intercept, which may imply additional considerations in the context of the spring constant calculation. The discussion also touches on the implications of plotting force against extension in relation to the expected linear relationship.

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Homework Statement


This is a lab and we have data from extending a crossbow and the force produced. We graphed the data (Force v Extension) and find a line of best fit. Just using the y intercept and slope my equation is y=1 + .61x. How do i use that to find the spring constant K?


Homework Equations



k=1/2(kx^2)
f=kx

The Attempt at a Solution


I used my equation and entered in some values for x to get the force. Then used them in the equation f=kx. But the values i obtain are all different based on x and therefore not a spring constant.

 
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shootyoup said:

Homework Equations



k=1/2(kx^2)
f=kx

you have F=kx, so if you plotted F against x you'd get a straight line passing through the origin. The gradient would be F/x which is?

In your case, the gradient of your line would still be F/x.
 
Here's an analogous situation which might help you.
Say I draw a linear fit between position(x) and time(t). What does the slope represent? Also notice, x=vt, where v is the (constant) speed.

Does this help?
 
Ahhh ok thanks guys, didn't know it was that simple
 

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