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Area Z represents the energy stored in the spring when it is stretched to a length L

I am trying to understand why this is so:

$$E = \frac{1}{2}kx^2$$

Here, $x$ is the vertical side of triangle $Y$. This way, $E = \frac{1}{2}x^2 = Area \ Y $ if and only if the slope of the line is $1$. What I am struggling to understand (or visualise to get an intuitive understanding) is how the factor $k$ compensates for this extra area when the slope does $not$ equal 1.

Any help will be greatly appreciated, thanks in advance.

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closed as unclear what you're asking by ACuriousMind, knzhou, CuriousOne, honeste_vivere, Gert Jun 23 '16 at 23:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Area of a triangle: 1/2*base*height. x is the base, kx is the height. $\endgroup$ – pentane Jun 20 '16 at 13:28
  • $\begingroup$ This appears to be a pure math question. Additionally, it is not clear what exactly you are not understanding about the formula for the area of a triangle. $\endgroup$ – ACuriousMind Jun 20 '16 at 13:36
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Precisely, the point is that the slope is not equal to 1. The ratio $\dfrac{F}{\Delta L}=k$. Therefore, $Z=\dfrac{1}{2}(x)*(\dfrac{x}{k})$ $= \dfrac{1}{2k}x^2$. Which looks wrong but is true because in the notation you are using, $x$ is not the elongation but is rather the force acting. So in a more familiar notation $E=\dfrac{1}{2k}F^2$. If you want to express the same formula in the terms of the elongation (let's call it $\Delta L = l$) then $E=\dfrac{1}{2k}F^2=\dfrac{1}{2k}(kl)^2 = \dfrac{1}{2}kl^2$.

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My explanation is as follows

Force on the spring in stretching it to length x can be written as

$F=k(x-L_0)$

where x is the displacement, $L_0$ is the initial length of the spring and k is spring constant. energy stored is

$dE=F.dx$

upon integration we will get

$E=\frac{1}{2}.k.(L-L_0)^2$

where $L$ is the final length of the spring. This is I think you already know.

now the length of triangle base in the graph = $W = F=k(L-L_0)$

length of triangle height = $(L-L_0)$

area of tringle = $\frac{1}{2}$.base.height

=$\frac{1}{2}k.(L-L_0)^2$

I think this will answer your query

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