Q:

Energy stored in series springs vs energy stored in parallel springs

I'm trying to understand why the energy stored in a set of series springs is different from the energy stored in parallel springs.
We know that the elastic potential energy stored in a spring system is as follows: $E=\frac{1}{2}k(\Delta l)^2$.
So imagine we have two identical springs each with a spring constant ($k$) of 85 Nm-1
In one system, they are in parallel, supporting a load of 15 N. In another, they are in series, also supporting 15 N.
So the combined spring constant in the parallel system is equal to $2k$, which is 170 Nm-1.
The combined spring constant in the series system is equal to $\frac{k}{2}$, which is 42.5 Nm-1.
Using the energy equation above, the energy stored in the springs is different for both systems, since $k$ is different and so is $\Delta l$.
I understand it from a mathematical point of view, but in terms of energy transfer, I don't understand why the elastic potential energy varies. Some energy is of course lost as thermal energy, but why don't both systems lose the same amount of energy or even gain the same amount of energy in the first place?

A:

We know that the elastic potential energy stored in a spring system is as follows: $E=\frac 12k\Delta  l$.

You are missing a power of 2 here: $E=\frac 12k(\Delta  l)^2$

Using the energy equation above, the energy stored in the springs is different for both systems, since $k$ is different and so is $\Delta l$.

If $k$ was different, then yes: the stored energy must also be different.
If $\Delta l$ was different, then yes: the stored energy must also be different.

If both are different, then you don't know. You can't conclude that the energy is different then. $k$ is halved, so what if $\Delta l$ is $\sqrt 2$ times as big? Then the same energy is stored.

Parallel springs help each other in carrying the weight. They can together in total exert double the force. Therefore you can consider this two-spring system as one equivalent spring with double the stiffness (since spring force $F=kx$ doubles with double stiffness $k$ if you compress it just as much):
$$k_{parallel}=2k$$
Since they each carry half of the load, they are each only compressed half as much as what they would have been alone:
$$\Delta l_{parallel}=\frac12\Delta l$$
Series springs have no help from each other. If you place a spring on a rigid table or on another spring makes no difference, since you look at the final situation (after compression is done). Therefore the top spring carries the whole load, and the second carries the whole load as well (plus the top spring's weight, but that's usually assumed mass-less). Each is compressed $\Delta l$, so in total the equivalent compression is double:
$$\Delta l_{series}=2\Delta l$$That is the same as having an equivalent only half as stiff spring: $$k_{series}=\frac12 k$$

Conclusion:

$E_{one\;spring}=\frac12k(\Delta l)^2$
$E_{parallel}=\frac12(2k)(½\Delta l)^2=\frac12\cdot2\cdot\frac14k(\Delta l)^2=\frac12E_{one\;spring}$
$E_{series}=\frac12(½k)(2\Delta l)^2=\frac12\cdot\frac12\cdot 4k(\Delta l)^2=2E_{one\;spring}$

Yes, the energies stored are indeed different. Because the displacements are different - you store energy in a spring by compressing/extending it, so different compression/extension means different energy stored.

