You probably mean that the voltage at node 4 is 5.7 V.
If there's current going right-to-left through the 150 Ohm resistor, then there must be a corresponding voltage difference between nodes 4 and 5 according to Ohm's law. So its not surprising that node 4 reads higher than 5 V.
If you are measuring 5.7 V at node 5, that would be very surprising, or your simulator is including some hidden behavior in the 5 V source that isn't shown in your drawing (i.e. an additional equivalent series resistance).
Reading the comments and reading the question again, it appears you're asking about a physical experiment and not a simulation experiment.
You don't say what kind of voltage source you have, and that's very important to the behavior.
If your voltage source is a battery, then the "classic" "non-ideal voltage source" model, with an ideal source and a resistor in series is pretty close to accurate. That would be the case where there is an additonal resistance that you haven't shown in your drawing, and as I explained above, when you push current back through that resistor, it produces a voltage across that resistor according to Ohm's law.
If your voltage source is a linear regulator (whether a "classic" or a low-dropout type) it depends exactly which model you use. But classic linear regulators typically completely fall out of regulation when you push reverse current through them. They no longer behave like voltage sources at all. Most likely they connect the output back to whatever the input supply was through a diode.But again it depends exactly what regulator you use --- some have reverse-bias protection that would make them look like an open-circuit instead.
If your voltage source is a benchtop supply or a wall-wart supply, then again it depends exactly what model supply and what its internal circuit is. Depending on the designer's expectations for how the supply is to be used, it could continue to look like a voltage source, it could shut itself down, or it could fall out of regulation and behave erratically when you push reverse current through it.