For capacitors, why does the dielectric need to be inserted at a small constant speed? Likely a very silly question.
I am aware that there is an attractive force from a charged parallel-plate capacitor in an open circuit without battery pulling the dielectric in, and that a decrease in energy will result upon insertion of the dielectric.
I am told that this difference in energy stems from the person's retarding force against the attractive force, but my question is why must this retarding force be applied? Why can't one just let the capacitor pull it in on its own? What would the energy transfers be in that case?
Thanks in advance!
 A: If it pulls it on his own, and you disregard friction it will come out on the other side, and so on and oscillate.
A: 
Why can't one just let the capacitor pull it in on its own?

In fact, that is what happens. But the effect on the energy stored in the capacitor will depend on if the capacitor is isolated or connected to a source when the dielectric is inserted.
Regardless of connection or disconnection, the capacitance will increase with the insertion of the dielectric (compared to a vacuum or air) this follows from
The capacitance of an isolated capacitor without the dielectric (e.g. air capacitor) is given by
$$C_{o}=\frac {\epsilon_{o}A}{d}$$
With the dielectric inserted,
$$C=\frac {\epsilon A}{d}$$
Since $\epsilon > \epsilon_{o}$, then $C>C_o$.
ISOLATED CAPACITOR
For an isolated capacitor, charge must be conserved. Then, given $Q=CV$ for charge to be conserved, if the capacitance increases, the voltage must decreases, thus the stored energy $E=\frac{1}{2}CV^2$ decreases since the energy varies linearly with capacitance and as the square of voltage. That means energy from the electric field of the capacitor was used to do work to "suck" in the dielectric, thus reducing the stored energy.
CAPACITOR CONNECTED TO BATTERY
On the other hand, if the capacitor is connected to the battery when the dielectric is inserted, instead of the charge being constant, the voltage across the capacitor remains constant at the same time as the capacitance increases. Then, since $Q=CV$ there is an increase in net charge $Q$ of each plate, which is due to work being done by the battery to supply the charge.
Now, since there is an increase in both capacitance for the same voltage, instead of the stored energy decreasing, it increases. The energy dumped into the capacitor equals the work done on the capacitor by the battery supplying the extra charge minus the work done by the capacitor in sucking in the dielectric.
Hope this helps.
