# Potential Energy of a Spring, Khan Academy **URL:** \ **Category:** Factual Questions\ **Created:** [June 15, 2021, 3:17am UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299 "2021-06-15T03:17:37Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![Someonewastaken](https://avatars.discourse-cdn.com/v4/letter/s/ba8739/32.png) [@Someonewastaken](https://boards.straightdope.com/u/Someonewastaken)\ **Post date:** [June 15, 2021, 3:17am UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/1 "2021-06-15T03:17:37Z")
[![](https://img.youtube.com/vi/eVl5zs6Lqy0/maxresdefault.jpg "Potential energy stored in a spring | Work and energy | Physics | Khan Academy") ](https://www.youtube.com/watch?v=eVl5zs6Lqy0) While watching this video, I had a question. I understand the calculus part and why he gets kx^2/2, but I don’t understand one other thing. If work=force times distance, than distance would be x and force would be kx, so my answer would be kx^2, not kx^2/2. I am pretty new to physics so I could be making a silly error, but why is my method wrong? ---
**Author:** ![Tabco](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tabco/32/7644_2.png) [@Tabco](https://boards.straightdope.com/u/Tabco)\ **Post date:** [June 15, 2021, 3:38am UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/2 "2021-06-15T03:38:18Z")
I want to say it’s because the force of the spring changes over the distance it moves. So you need to develop an integral to represent the total work done. In this case, the integral of kx from zero to x with respect to x. That integral will give you 1/2 kx^2. That’s my guess, anyway. ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [June 15, 2021, 4:22am UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/3 "2021-06-15T04:22:22Z")
@Tabco more or less got it. Work is formally defined as a line integral of Force with respect to displacement along the curve. For a spring this works out to the integral of F\*v dt, where v is the velocity and this ends up being (1/2)kx^2. In the video, it is explained that the force ends up being ~ kx/2 (the area under the curve) and not directly k\*x by itself, which is more or less equivalent to working out the integral. ---
**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\ **Post date:** [June 15, 2021, 5:28am UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/4 "2021-06-15T05:28:36Z")
The area under the curve is a triangle, which has area base\*height/2. The base is x (the distance over which we travel) and the height is kx (the force at maximum extension). So the area is just x\*kx/2, or kx^2/2. Same result as the integral, of course. ---
**Author:** ![Someonewastaken](https://avatars.discourse-cdn.com/v4/letter/s/ba8739/32.png) [@Someonewastaken](https://boards.straightdope.com/u/Someonewastaken)\ **Post date:** [June 15, 2021, 5:41am UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/5 "2021-06-15T05:41:42Z")
Ohhhhh! Thank you all! ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [June 15, 2021, 3:30pm UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/6 "2021-06-15T15:30:44Z")
To put it in a different way and connect it with another piece of elementary physics. If x is the current extension of the spring the current force on the spring is kx, but at half the extension the force was kx/2 and at no extension the force was zero. And since the force is linear with respect to the distance, the average over the distance x is kx/2 and you can multiply that average force by the distance and get the work. You have something similar when looking at situations with constant acceleration. If you start at zero speed and accelerate at a constant rate to a speed v, your speed was v/2 halfway (with regards to time), your average speed is v/2 and your distance is vt/2. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [June 15, 2021, 3:41pm UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/7 "2021-06-15T15:41:25Z")
> [@naita](#): > > You have something similar when looking at situations with constant acceleration. If you start at zero speed and accelerate at a constant rate to a speed v, your speed was v/2 halfway (with regards to time), your average speed is v/2 and your distance is vt/2. Which is, of course, mathematically the same integral. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [June 15, 2021, 4:52pm UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/8 "2021-06-15T16:52:16Z")
Indeed, but helpful to build a better understanding for someone new to both manipulating physics formulae in general and particularly to doing so using integrals. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [June 15, 2021, 5:49pm UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/9 "2021-06-15T17:49:47Z")
Oh, sure, I wasn’t criticizing your description (which is in fact the way I usually present it to non-calculus students, too). I meant to reinforce it, by pointing out that the two situations look similar because they are similar. ---
**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\ **Post date:** [June 26, 2021, 8:55am UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/10 "2021-06-26T08:55:48Z")
It was the very basic intergration process known as “averaging”. Area of a triangle ? use the average height… ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [June 26, 2021, 7:29pm UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/11 "2021-06-26T19:29:39Z")
I’m not sure what you’re trying to say. But here’s is something I chose to leave out previously. As I wrote in the first reply, in the OPs question,and in a situation with constant acceleration the force or speed to use is the average. But if you don’t understand why and how that average is found, it’s easy to misapply averages. An understanding of seeing distance as the area under the speed/time graph can be useful here, even if you don’t know that that is a a form of integration. Without that understanding you can get trapped by questions such as “You drive 1 lap at 12 mph and 1 lap at 60 mph. What was your average speed?” or the infamous “You drive one lap at 50 mph. How fast do you have to drive a second lap to make the average speed 100 mph?” ---
**Author:** ![Limmin](https://avatars.discourse-cdn.com/v4/letter/l/f04885/32.png) [@Limmin](https://boards.straightdope.com/u/Limmin)\ **Post date:** [June 26, 2021, 11:09pm UTC](https://boards.straightdope.com/t/potential-energy-of-a-spring-khan-academy/944299/12 "2021-06-26T23:09:09Z")
Great explanations here. The assumptions being made about the triangle or integral for the work of the spring is that the spring is _linear_, meaning the curve makes an actual triangle.