"Hot-knife"/Thermal cutter power supply design
You do not need to mess with your power outlets to build a thermal knife/cutter. Thermal knives operate by using the Joule effect to provide enough energy to the material to cut in order to heat it up past its fusion point until it melts (or up to its flash point until it burns).

You need Joule dissipation power:
$$P_J=RI^2$$

A high resistance wire will give you more heating power, but will require a higher voltage to provide the same amount of current. Looking at the above formula, if you have reached the maximum voltage of your supply, stop there, as current is more important than resistance. 

Resistance of a wire :
$$R=\rho*L/A$$
Where rho is the resistivity of the conducting material, L the length, and A the cross section area. L can be increased by winding the wire around the material, and A decreased using higher AWG gauge wire. Don't forget to remove the insulation of the wire, as it represents a thermal resistance that will decrease the power making its way out of the wire.

Your first requirement is the required amount of energy to cut the rope. This energy is
$$E_{melt}=m\Delta T C+mH_{fusion}$$
Where the first part is the amount of energy to heat the material up to its fusion/flash point (m is the mass of that material, $\Delta T$ is the difference in temperature, C the specific thermal capacity), and the second part the amount of energy to complete the fusion/combustion process ($H_{fusion} is the specific heat/enthalpy of fusion$).
Note that if the thermal resistance of the material to cut is not negligible (as we can expect from a rope), this formula will give you an upper bound as the heating power will not diffuse throughout the material, and heat a specific spot instead. 
This assumes no losses to the environment though, I advise proper thermal insulation around the cutter, and to use thermal paste between the cutter and the material.

Fire up Excel, and figure out what you need. I think a 2A DC wall wart could do the trick for you with suitable wire. If you use a transformer for higher voltages and/or currents, do not forget that Joule heating is in function of RMS current.