Work to stretch a spring question (nvm, figured it out)

[gibberingmouther]

<Moderator's note: Moved from a technical forum and thus no template.>

NEVERMIND: yeah i get it now. y = k*X literally at x = X. don't know why i had trouble with that

so i see how to derive .5 * k * X^2 (work done to move a spring from rest to a length X) from calculus. but i was looking at the demonstration using a graph and i am confused about it. so the area of under the graph of y = k * x at spring length X is equal to .5 * k * X^2. you can verify this with calculus, but i don't see how it works by geometry. X ^2 + ycomponentoftriangle^2 = k^2*X^2 -> ycomponentoftriangle = X*(k^2 - 1)^.5

here's a picture of what I'm talking about: http://www.batesville.k12.in.us/Physics/PhyNet/Mechanics/Energy/spring_work_2.htm

using pythagorean's theorem i got a different result for the value of the y component of the triangle. i don't understand how it is equal to k*X (or k*x in the link) using geometry.

 

[BvU]

What is the work equal to in your relevant equation ?
You really should use the homework forum with a useful template for this. See guidelineshttps://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/

Homework Statement

Homework Equations

Work $W$ if taking small steps $\Delta x$ is $W = \sum F \cdot \Delta x\$ with $\Delta x$ so small that $F$ does not change appreciably. In the limit of $\Delta x \downarrow 0$ you get the integral $W = \int F(x)\, dx$ which is the area under the curve (straight line) $F(x)$. Nothing to do with Pythagoras or a y component.

The Attempt at a Solution

 

[gibberingmouther]

What is the work equal to in your relevant equation ?
You really should use the homework forum with a useful template for this. See guidelineshttps://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/

Homework Statement

Homework Equations

Work $W$ if taking small steps $\Delta x$ is $W = \sum F \cdot \Delta x\$ with $\Delta x$ so small that $F$ does not change appreciably. In the limit of $\Delta x \downarrow 0$ you get the integral $W = \int F(x)\, dx$ which is the area under the curve (straight line) $F(x)$. Nothing to do with Pythagoras or a y component.

The Attempt at a Solution

it's not homework though I'm using a textbook (i took physics 1 while working on my associate's but had to withdraw because i got overloaded). i was just trying to understand the geometrical reasoning behind the formula for the work done by stretching a spring. i.e. W = .5 * k * X^2 where k is the spring constant and X is the length the spring is being stretched. i couldn't figure out why the vertical side of the triangle (see my link) was k*X. then i realized that i had overlooked the obvious: if x = X, then y = k*X, algebraically speaking.

 

[BvU]

Yes.
Is my story about the area understandable ?

 

[gibberingmouther]

Yes.
Is my story about the area understandable ?

yes. i took Calc I and II for my degree and i still remember how to do derivation and integration. i can see how to derive the work equation using calculus, but my textbook also had a geometrical demonstration that i initially couldn't understand. i know it's the same thing, basically, but i like to follow all the demonstrations in my textbook. if my textbook doesn't explain where an equation comes from, i'll try to figure it out myself or more likely look on the internet for a demonstration.

 

[BvU]

I added a link to #5