# Energy Stored in Spring [closed]

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.

## closed as unclear what you're asking by ACuriousMind♦, knzhou, CuriousOne, honeste_vivere, GertJun 23 '16 at 23:48

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• Area of a triangle: 1/2*base*height. x is the base, kx is the height. – pentane Jun 20 '16 at 13:28
• 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. – ACuriousMind Jun 20 '16 at 13:36

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 $E=\dfrac{1}{2k}F^2=\dfrac{1}{2k}(kl)^2 = \dfrac{1}{2}kl^2$.

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

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