# Understanding the concept of energy in a vertical spring system

I have a concept I don't think I get well. I would like to clear out a couple of things. Consider the following system:

The spring on the left is the one on the right after extension from equilibrium. I tried calculating the spring's total energy on the right case with respect to h=0, considering it has 0 velocity there. First I did it with respect to h=0 where the spring isn't extended at all, and then I did it with respect to the new equilibrium position, thus considering gravitational potential energy in the spring energy itself. That's what I got:

My question is, what is the physical meaning of this C? I expected to get the same energy both ways. Why did I get a larger energy E2? how do I equate them and why?

*This isn't homework, I am just baffled by this concept and why I always get different energies in different ways.

• Care to explain what is the problem in my way? that might help me. Commented Feb 28, 2020 at 15:51
• Why would it be zero? Sorry if it's clear, I just don't see the problem. I calculated both E1 and E2 at the same position, where the body has maximum spring potential energy (it stopped moving). Commented Feb 28, 2020 at 16:40
• It oscillates up and down with x0 being the equilibrium. On the right picture, you see it extended a distance d-x0 from equilibrium, to a point in which it is at rest and about to go back up. Commented Feb 28, 2020 at 20:56
• Darkenin, my previous response is in error. I will post something from an AP Physics C textbook that helps clear the issue up. Commented Feb 29, 2020 at 0:41

$$E_2$$ is greater than $$E_1$$ because when you calculate $$E_2$$ you assume the weight is initially at rest at $$h=0$$ whereas when you calculate $$E_1$$ you assume the weight is initially at rest at $$h=x_0$$.

If you release the weight from rest at $$h=0$$ then at extension $$h$$ the net force on the weight is $$F(h)=mg-kh$$. If we integrate $$F(h)$$ from $$h=0$$ to $$h=x$$ we find the kinetic energy of the weight as it passes through the point $$h=x$$:

$$KE(x)=mgx - \frac 1 2 kx^2$$

When $$x=x_0=\frac {mg}{k}$$ we have

$$KE(x_0) = \frac {(mg)^2}{k} - \frac {(mg)^2}{2k} = \frac {(mg)^2}{2k} = C$$

So $$C$$ is the kinetic energy that the weight has at $$h=x_0$$ if it released from rest at $$h=0$$. In other words, $$C$$ is the energy that you have to remove from the system in order to bring the weight to rest at its equilibrium position $$h=x_0$$, which is your baseline for calculating $$E_1$$.

Two problems: The value of a potential energy depends on your choice of a reference point (where the energy is zero). Your E1 puts the energy at zero when d = 0. Your E2 puts the energy at zero when d = Xo. They are not expected to be the same. Also you included the gravitational energy in E1 but did not include it in E2.