Braking with less resistance
R = V^2 / P = 36^2 / 500 = ~2.5 Ohm


It looks like you've assumed it's a 36v motor. It will generate about that when spun at its rated speed, so neglecting losses, this calculation is correct.

If you use very much less than this resistance, then two things will happen.

1) At its rated speed, at 36v output, much more current will flow than you expected, so the braking torque will be higher than you'd planned, as will the braking power.

2) The motor windings and the braking resistor carry the same current. If the braking resistor is very small, then the braking power will be dissipated as heat in the motor. If you just have a spinning motor by itself and stop it, that's not too much heat. If you are stopping a laden vehicle running down a hill, then that will probably overheat your motor. That's why you'll often see the loco for a mountain railway has a rack of air-cooled resistors on the roof, that's where the braking power gets dissipated, not in the motor windings.

Braking with a resistor is dynamic, that is, the braking torque drops as the speed drops. If you want steady braking torque, you need a controllable load. As the motor has an inherent inductance, and as there's no need for clean supplies, this can be done as simply as using a FET and a freewheel diode to chop down to a resistive load, a crude buck converter. Control the switching to control the braking torque as required.