Conservation of energy in a solenoid I found a website this afternoon which stated that for a given solenoid, the force it exerts on an iron bar can be modeled as a conservation of energy problem.
The page gives a method to find $\Delta u_{solenoid}$. It then says that the force exerted on the bar can be modeled as
$$F= \frac{\Delta u_{solenoid}}{\text{length of solenoid}}$$
This is a problem for me because I need the work done on the bar.
For a bar of identical length to the solenoid, if I try to utilize 
\begin{align}
W &= |F|\cdot|s| *\text{cos }\theta\\
&= |\frac{\Delta u_{solenoid}}{\text{length of solenoid}}| \cdot|\text{length of solenoid}| *\text{cos }\theta \\
&=|\Delta u_{solenoid}|*\text{cos }\theta
\end{align}
Then I'm stuck with something dependent on $\theta$. Intuitively, and experimentally, I know that magnets pull on iron which suggests to me that in this scenario $\theta$ is always $0$. (Both poles of a magnet pull iron, etc.)
However, this doesn't seem like the "proper" way to do it, and I was wondering if anyone could explain analytically what $\theta$ has to be and why. 
Or, considering the not-improbable scenario in which I'm approaching this problem all wrong, what would be the correct way to get $W$ from my $\Delta u_{solenoid}$?
Edit:
I should have explained a bit better. The way that I view this
$$ 0=\Delta KE + \Delta u$$
$$\Delta u = -\Delta KE$$
Which suggests to me that when the iron bar enters the solenoid, $KE$ is reduced and negative work is done on the bar. 
Experimentally, this isn't the case, however, and I'm trying to determine what's going on. I suspect that it has to do with a change in potential energy of my power source, but I'm not familiar enough with electrical and magnetic potentials to say.
 A: I think I've worked out a solution after sketching some things out.

If we take the iron bar to be motionless and the solenoid to be moving, then the force on the solenoid is $$F_{solenoid}=\frac{\Delta u_{solenoid}}{-L}$$ From Newton's Third Law, we know that there's an antiparallel force of the same magnitude acting on the bar. $$F_{solenoid}=-F_{bar}$$
Now if we consider the solenoid to be stationary and the iron bar to be in motion, then 
$$W_{bar} = |F_{bar}| \cdot |s| * \text{cos } \theta$$
$$W_{bar} = |-F_{solenoid}| \cdot |L| * \text{cos } 0^{\circ}$$
$$W_{bar} = |\frac{\Delta u_{solenoid}}{L}| \cdot |L|$$
$$W_{bar} = |\Delta u_{solenoid}|$$
Which answers my question. 
I think I was accidentally confusing everything by treating the $KE$ of my solenoid as the $KE$ of the iron bar.
Edit:
To satisfy Conservation of Energy:
If one has an inductor, there must be a power source somewhere with $E=u_{battery}$. For the solenoid/battery system, which is stationary if the bar is moving:
In general:
$$W_{other}=\Delta KE_{solenoid} + \Delta u$$
$$-W_{bar} - W_{heating} = \Delta KE_{solenoid} + \Delta u_{field} + \Delta u_{battery}$$
$$-W_{bar} - W_{heating} = (0) + \Delta u_{field} + \Delta u_{battery}$$
and specifically for our situation:
$$-|\Delta u_{field}| - W_{heating} =\Delta u_{field} + \Delta u_{battery}$$
$$-\Delta u_{field} -|\Delta u_{field}| - W_{heating} =\Delta u_{battery}$$
And because our $\Delta u_{field}>0$
$$-2\Delta u_{field} - W_{heating} =\Delta u_{battery}$$
$$\Delta u_{battery}= -2\Delta u_{field} - W_{heating} $$
Which shows that our example does indeed follow the Law of Conservation of Energy. We're not getting energy from nothing; it must come from the power source.
