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**Where does 1/2 come from in 1/2kx^2??**

Sorry is this sounds like a silly question, but i was just wondering where the 1/2 in PE = 1/2kx^2 comes from. If F=kx, then wouldnt it be kx*x = kx^2 ?

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Sorry is this sounds like a silly question, but i was just wondering where the 1/2 in PE = 1/2kx^2 comes from. If F=kx, then wouldnt it be kx*x = kx^2 ?

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arildno

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[tex]PE=\int_{0}^{x}kx'dx'=\frac{1}{2}kx^{2}[/tex]

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errr....I havnt taken calculas yet, so i have no idea what that means, lol.

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Doc Al

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Even without calculus you should be able to see where the half comes in. In the equation Work = Force * Distance, the force is changing so you can't just put F = kx, since that's only true when the spring is fully stretched. Instead, use theArmoSkater87 said:errr....I havnt taken calculas yet, so i have no idea what that means, lol.

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[tex] F = kx [/tex]

now you find the anit - derivative, do you know what that means?

[tex] P_e = \frac {1}{2}kx^2 [/tex]

if you still dont understand, pm me.

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chroot

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It doesn't take much calculus to perform an integration, Nenand?

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chroot said:It doesn't take much calculus to perform an integration, Nenand?

- Warren

chill, chill, chroot, hmmm,..., Nenand, chroot ???

he was only saying integrating a constant multiplied by x is one of them elementary integranda...

marlon

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The "area under the curve" is the "work done by that force".

Since F(x) is linear, you are finding the

So, (Work done by F)=(1/2)(x)(-kx).

Since that force is conservative, the potential energy is minus the work done by that force.

So, U=-(-(1/2)kx^2)=(1/2)kx^2.

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integration is calculus. And I wasnt shure if he had any experience with polynomial functions or calculus.chroot said:It doesn't take much calculus to perform an integration, Nenand?

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chroot

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I'm aware.Nenad said:integration is calculus.

He said he didn't.And I wasnt shure if he had any experience with polynomial functions or calculus.

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russ_watters

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To get you started, an integral is the area under a curve and the derivative is the slope of the curve. Its more complicated than that, of course...ArmoSkater87 said:

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NateTG

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You should already know that the force that a spring exerts is proportional to the displacement (i.e. [tex]F=kx[/tex]), and be familiar with the work-energy equation for a constant force ([tex]W=F d[/tex]).ArmoSkater87 said:Sorry is this sounds like a silly question, but i was just wondering where the 1/2 in PE = 1/2kx^2 comes from. If F=kx, then wouldnt it be kx*x = kx^2 ?

Now, we're not entirely sure about what the work that is done by the spring is, but we can approximate it. For example, we could cut the path of the spring into lots of little pieces, and approximate the work done on each of the peices of the path by picking the smallest force that the spring exerts on that segment to calculate the work for that segment. It turns out that it's possible to show that the total of these sengments tends toward [tex]\frac{1}{2}kx^2[/tex] as the segments get smaller and smaller.

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Thanks for the tip.russ_watters said:To get you started, an integral is the area under a curve and the derivative is the slope of the curve. Its more complicated than that, of course...

Yes, I already know that...NateTG said:You should already know that the force that a spring exerts is proportional to the displacement (i.e. [tex]F=kx[/tex]), and be familiar with the work-energy equation for a constant force ([tex]W=F d[/tex]).

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