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

  • Context: High School 
  • Thread starter Thread starter ArmoSkater87
  • Start date Start date
Click For Summary

Discussion Overview

The discussion centers around the origin of the factor of 1/2 in the potential energy formula PE = 1/2 kx^2, specifically questioning how this factor arises in the context of spring force and work done. The scope includes conceptual understanding and basic reasoning related to physics principles, particularly in relation to work and energy in springs.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the presence of 1/2 in the equation, suggesting that if F = kx, then it should simply be kx^2.
  • Another participant explains that the work done must be represented as an integral, leading to the conclusion that PE = 1/2 kx^2.
  • Some participants note that the force exerted by the spring is not constant and that the average force should be used, which is kx/2, leading to the same conclusion.
  • One participant mentions that understanding the concept of integration is not necessary to grasp the explanation of the 1/2 factor.
  • Another participant describes the work done as the area under the curve of a linear force vs. displacement graph, which is a triangle, thus yielding the factor of 1/2.
  • There are multiple references to calculus concepts like integration and derivatives, with some participants expressing confusion about these terms.
  • A later reply suggests approximating the work done by considering small segments of the spring's displacement, leading to the conclusion that the total work approaches 1/2 kx^2 as segments become infinitesimally small.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the mathematical concepts involved, particularly integration. While some explanations are accepted, there is no consensus on the necessity of calculus for understanding the derivation of the 1/2 factor, and the discussion remains somewhat unresolved regarding the clarity of these concepts for all participants.

Contextual Notes

Some participants indicate a lack of familiarity with calculus, which may limit their understanding of the explanations provided. There is also uncertainty about the exact nature of the work done by the spring and how it can be approximated without calculus.

ArmoSkater87
Messages
253
Reaction score
0
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 wouldn't it be kx*x = kx^2 ?
 
Physics news on Phys.org
In this case the (force*distance) formula must be represented as an integral, for example:

[tex]PE=\int_{0}^{x}kx'dx'=\frac{1}{2}kx^{2}[/tex]
 
errr...I havnt taken calculas yet, so i have no idea what that means, lol.
 
force is not constant

ArmoSkater87 said:
errr...I havnt taken calculas yet, so i have no idea what that means, lol.
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 the average force: The force varies uniformly from 0 to kx, so the average force is kx/2. So Work = (kx/2)*(x) = 1/2 k x^2. Make sense?
 
basically, what arildno did was integrate. You don't need much calculus to understand. If you have
[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 don't understand, pm me.
 
It doesn't take much calculus to perform an integration, Nenand?

- Warren
 
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
 
Plot force F(x) vs displacement x.
The "area under the curve" is the "work done by that force".

Since F(x) is linear, you are finding the area of a triangle with base (x) and height (-kx).
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.
 
chroot said:
It doesn't take much calculus to perform an integration, Nenand?

- Warren

integration is calculus. And I wasnt shure if he had any experience with polynomial functions or calculus.
 
  • #10
Nenad said:
integration is calculus.
I'm aware.
And I wasnt shure if he had any experience with polynomial functions or calculus.
He said he didn't.

- Warren
 
  • #11
Oh wow, that makes sense, at least the explanations without the use of "intergration" or "derivative" and that calculus stuff. I have no idea what those words mean, although I've heard them many times. I'm going to take BC calc when school starts. Anyways, thanks a lot for the responses. :)
 
  • #12
ArmoSkater87 said:
Oh wow, that makes sense, at least the explanations without the use of "intergration" or "derivative" and that calculus stuff. I have no idea what those words mean, although I've heard them many times. I'm going to take BC calc when school starts. Anyways, thanks a lot for the responses. :)
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...
 
  • #13
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 wouldn't it be kx*x = kx^2 ?

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]).

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.
 
  • #14
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...

Thanks for the tip. :smile:

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]).

Yes, I already know that...
 

Similar threads

Undergrad Work-Energy Theorem for Moving Block and Spring System?
  • · Replies 4 ·
Replies
4
Views
3K
Finding a formula for displacement of a mass on a spring using v.
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad A Continuous Solution for Mass/Spring w/ Friction
  • · Replies 17 ·
Replies
17
Views
3K
Undergrad What Happens to the Excess Energy in a Compressed Spring?
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Change of coordinates in a potential energy field
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Help with IB Physics IA about bungee jumping....
  • · Replies 16 ·
Replies
16
Views
5K
High School Where does the 1/2 in 1/2 at^2 come from?
  • · Replies 48 ·
2
Replies
48
Views
6K
Problem of spring block system: force vs conservation of energy
  • · Replies 20 ·
Replies
20
Views
4K
Undergrad Why the work done is coming out to be positive?
  • · Replies 9 ·
Replies
9
Views
2K
Is Equating Energies a Better Solution for Finding Extension Under Spring Force?
  • · Replies 4 ·
Replies
4
Views
1K