# Understanding the Work Energy Theorem for a Spring

1. May 17, 2015

### mrpolar

Hi,

I looked around for hours but it seems like I'm the only one who finds it confusing.
I understand the concept of potential energy and work, but have a problem with the equations.

Here is what I don't understand:

The work energy theorem states that K2-K1=W.
W = ∫Fdx from evaluated between x2 and 1.
The Force done by a spring is -kx, so when only the spring acts on the body the equation should be:
K2-K1=-½kx22 + ½kx12
K2+½kx22=K1+½kx12

But now, since W=-U(x) (potential energy) the equation turns out to be:
K2-½kx22=K1-½kx12
For the total energy balance (kinetic + potential). But the lectures and examples I saw, and also the logical thing as I understand it, is that the equation should be K2+½kx22=K1+½kx12 for the kinetic + potential energy balance.
Why do I get the signs wrong?

Here's an example for a question which made me really confused:
http://oyc.yale.edu/physics/phys-200/lecture-5#ch3 (minute 40:30).
In short, a mass is hanged from a spring attached the ceiling, so the mass has 2 forces acting on it: gravity and the spring.
In order to find the kinetic energy - potential energy equation, we have ΔKinetic_Energy = the negative of the work (-∫Fdx) of the total forces.
Now I know that gravity acts downwards, so the force is -mg, and the spring's force is always -kx, so the equation should look something like this:

K2-K1 = -∫F(gravity)dx - ∫F(spring)dx evaluated between x2 and 1. The negative signs before integrals is because we want to find the potential energy which is the negative of the work done.
K2-K1 = -∫-mgdx - ∫-kxdx = ∫mgdx + ∫kxdx
K2-K1 = mgx2 +½kx22 - mgx1 +½kx12

K2-mgx2-½kx22 = K1-mgx1-½kx12
which is the total energy equation.

In class, they had the potential energy signs all opposite like this: K2+mgx2+½kx22 = K1+mgx1+½kx12 which of course looks more sensible and logical.

The problem is that I can't see where I got the maths wrong. I follow the rules of and get it all messed up time and time again.

Please, help me out here, this is driving me nuts!
Ben.

2. May 17, 2015

### nasu

Why do you integrate from x2 to x1 when calculating the work?

3. May 17, 2015

### mrpolar

I evaluate the overall work done between the starting point (x1) to the end point (x2), so I integrate from x2 to x1 (the value at x2 minus the value at x1 gives me the work done between). I got something wrong?

4. May 18, 2015

### nasu

Yes, you need to integrate from x1 to x2.

5. May 18, 2015

### mrpolar

When I say that I integrate from x2 to x1 I mean that I set x2 as the upper bound and x1 as the lower bound of the integral.
You mean to set x1 as the upper bound and x2 as the lower bound in order to calculate the work?
I thoughts you do it in order to find the potential energy (hence the minus sign before the integral which is the same as reversing the integration order).

6. May 18, 2015

### nasu

Then you use the words in an unusual way. Integral from a to b means that a is the lover limit and b is the upper limit.
See here, for example:
http://tutorial.math.lamar.edu/Classes/CalcI/DefnofDefiniteIntegral.aspx

What you get after integration and regrouping the terms is the right conservation of energy equation. The potential energy at x1 is 1/2kx1^2, given that the zero potential energy is at x=0.
The definition of potential energy tells you that the work is the negative of the change in potential energy.
So the change in PE from x1 to x2 is the negative of the work from x1 to x2, This is
Δ PE=1/2k*x2^2-1/2k*x1^2
So if you want to use conservation of energy rather than work-energy theorem, you have
ΔKE+ΔPE=0
which will give you again the correct equation
KE1+1/2k*x1^2=KE2+1/2k*x2^2

Last edited: May 18, 2015
7. May 18, 2015

### mrpolar

Hi, thank you again, you are really helpful. Sorry the wrong terminology.
So, if I got it right, for a mass hanged by a spring from a ceiling,
ΣF =-mg-kx (the spring's equalibrium = 0 potential energy, so gravity can take values of minus potential energy whenever the mass is beneath the equalibrium. is it ok?)
ΔKE = ΔW = ∫-mg-kx dx = -mgx2 -½kx22 + mg1 + ½kx12
KE2 + mgx2 + ½kx22 = KE1 + mg1 + ½kx12

First question: those equations are right? will work in any x point I define no matter if the mass is going up or down, beanth or above equalibrium, right?
My second question, I know that this equation is essentially KE2 + PE2 = KE1 + PE1[/SUB
but the righthand side of the equation is the kinetic energy at x1 plus the x1 part of the work equation, not the potential energy (I didn't multiply that part by (-1) the same as I did to the x2 part of the ΔW by moving it to the right-hand side of the equation). What's going on here?

8. May 19, 2015

### nasu

Yes, it is the potential energy. As it is the "x1 part" of the work-energy equation.
Your W is -ΔPE =-(PE2-PE1)=-PE2 +PE1.
So the x1 part is PE1. And the x2 part is PE2.
But it's no point to go keep going between work-energy and conservation of energy. Decide which one you want to use and stay with it for the given problem.

9. May 21, 2015

### mrpolar

Thank you so much for helping me out here. I finally 100% got it.
Time to figure out conservation of energy in 2 dimensions :P

Thanks again.