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

  • Thread starter ArmoSkater87
  • Start date
In summary: You don't need much calculus to understand. If you have a linear function, like F=kx, then you can integrate to find the total area under the curve. However, for a nonlinear function, like F(x), you need to use calculus to find the slope of the function and find the area under the curve.
  • #1
ArmoSkater87
253
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
  • #2
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]
 
  • #3
errr...I havnt taken calculas yet, so i have no idea what that means, lol.
 
  • #4
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?
 
  • #5
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.
 
  • #6
It doesn't take much calculus to perform an integration, Nenand?

- Warren
 
  • #7
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
 
  • #8
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.
 
  • #9
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...
 

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

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

The 1/2 in the equation 1/2kx^2 represents the coefficient of the quadratic term, which is a result of the mathematical derivation of the potential energy stored in a spring. It is derived from the integral of the force applied by the spring over a displacement x.

2. What does the k in the equation represent?

The k in the equation 1/2kx^2 represents the spring constant, which is a measure of the stiffness of a spring. It is a characteristic property of a spring and is determined by its material and dimensions.

3. How is the equation 1/2kx^2 related to Hooke's Law?

The equation 1/2kx^2 is a mathematical representation of Hooke's Law, which states that the force applied by a spring is directly proportional to the displacement of the spring from its equilibrium position. The 1/2kx^2 term represents the potential energy stored in the spring due to this displacement.

4. Can the 1/2 in the equation be changed to a different value?

No, the 1/2 in the equation 1/2kx^2 is a constant and cannot be changed. It is a fundamental part of the equation and is derived from the integral of the force applied by the spring over a displacement x.

5. Why is the 1/2 in the equation important?

The 1/2 in the equation 1/2kx^2 is important because it represents the relationship between the force applied by the spring and the displacement of the spring from its equilibrium position. It is a crucial term in calculating the potential energy stored in a spring and is essential in understanding the behavior of springs in various physical systems.

Similar threads

I A Continuous Solution for Mass/Spring w/ Friction
    • Classical Physics
Replies
17
Views
1K
B Where does the 1/2 in 1/2 at^2 come from?
    • Classical Physics
2
Replies
48
Views
2K
B How Does Phase Change Affect the Equation of a Stationary Wave?
    • Classical Physics
Replies
12
Views
1K
Problem of spring block system: force vs conservation of energy
    • Introductory Physics Homework Help
Replies
20
Views
2K
Is Equating Energies a Better Solution for Finding Extension Under Spring Force?
    • Introductory Physics Homework Help
Replies
4
Views
856
Spring with mass hangs from the ceiling
    • Introductory Physics Homework Help
Replies
22
Views
500
Spring constant formlula for Force and Work Done
    • Introductory Physics Homework Help
Replies
12
Views
782
Force components for mass attached to two springs
    • Introductory Physics Homework Help
Replies
15
Views
290
B Understanding Phase Terminology in Wave Functions
    • Classical Physics
Replies
9
Views
865
2 masses connected by spring, one is pulled, how much does the spring stretch?
    • Introductory Physics Homework Help
Replies
3
Views
656

Hot Threads

Recent Insights

Back
Top