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OpenStudy (anonymous):

In the most detail you can possibly explain, how would I find the 'K' or the spring constant of a slope where the 'y' axis is labeled as Newtons, and the 'x' axis labeled as the change in length?

OpenStudy (roadjester):

just find the slope of the line

OpenStudy (anonymous):

I probably should have taken more time to explain this, but I know the equation in this case is F=K times the change in length, but with the multiple numbers associated with this graph, I'm not sure how to go about completing the equation

OpenStudy (roadjester):

Could you possibly scan it?

OpenStudy (anonymous):

Yeah I'll take a picture, one second

OpenStudy (roadjester):

Just keep in mind that for oscillatory motion, the spring constant k is linear.

OpenStudy (roadjester):

Look on page 2 of the pdf.

OpenStudy (anonymous):

bleb

OpenStudy (roadjester):

Okay wha...the spring force is a restorative force, that graph should have a negative slope...

OpenStudy (anonymous):

Well fack. Maybe I've misunderstood this. I'll get help with it in class tomorrow.

OpenStudy (roadjester):

What physics class is this?

OpenStudy (anonymous):

Physics 1

OpenStudy (roadjester):

High school or college?

OpenStudy (anonymous):

High school

OpenStudy (roadjester):

Okay, thanks. Now, do you know what a "restoring force" is?

OpenStudy (anonymous):

yeah

OpenStudy (roadjester):

Okay, so a spring force is a restoring force. When you don't touch it, a spring is in equillibrium and at rest. When you apply a force and displace it a certain length, you don't really notice the restoring force. But the more you pull the harder it is to pull right? That's because the "restoring force" is "fighting" against you. So a spring force should be negative.

OpenStudy (roadjester):

If you look at the pdf I attached, on the second page is what a spring force-displacement graph looks like.

OpenStudy (anonymous):

Oh, yes that does make sense, I can't find the PDF attached though lol.

OpenStudy (roadjester):

OpenStudy (anonymous):

Alright, I'll have a look and continue on with my works. Thanks a lot. c:

OpenStudy (roadjester):

No prob. oh and @Unitystep I wouldn't recommend trying to understand those notes unless you have at least a basic understanding of calculus. That's from my own physic class way back when and it was an calculus based physics course (college level)

OpenStudy (roadjester):

You can try, but don't hurt yourself trying

OpenStudy (anonymous):

I take a pre-calculus class so I should be fine lol.

OpenStudy (roadjester):

Up to you. Just don't forget to close the question. =) Good luck!

OpenStudy (mrnood):

@Unitystep This is not a calculus issue Spring force is a linear relationship The graph detail shows that (over the elastic portion of the spring extension) the relationship of force to extension is F=-kx For any straight line graph of the form y=mx+c the value of m is the slope of the graph. So - if you have a graph with data points on it (at least 2 points are required) then calculate the slope - that is the spring constant. The slope is calculated by (change in Y)/(change in X) For a straight line you can get these values from any two points on the graph. (you would need differential calculus to find the slope at a given point IF the line was not a straight line...)

OpenStudy (roadjester):

@MrNood The calculus I mentioned wasn't for this problem. It was for interpretation of the notes I attached.

OpenStudy (mrnood):

I see that now. The OP's graph is a clear straight line an d the data is precisely linear. Just a question of calculating the slope...

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