Need filter design to increase DC motor brush life
Without a capacitor after the bridge rectifier you are actually applying a folded sine-wave to the motor, not a true DC voltage.



As such the voltage across the motor is changing from zero volts to the peak voltage at twice the line-frequency. Worse the RMS voltage of 120V peaks at 140 * 1.414 = ~170V, which is higher than the rated voltage of the motor.

For proper operation you need to smooth out the DC voltage using a sizable capacitor rated for at least 200V across the output of the rectifier.



Since the load draws current between the peaks from the regulator the capacitor needs to be large enough to supply that current between them without dropping too much voltage. The size of the capacitor therefor depends on how much current is drawn by the load.

With full-wave rectification, the capacitor must supply current to the regulator for half a cycle (t = 8.3ms @ 60Hz,  t = 10ms @ 50Hz).

As such, the Charge Q (Coulombs) removed from the capacitor is Q=I*t, where I is current and t is time.

You need to limit the droop during that period to some voltage drop, say 10%, or 12V for a 120V.

Since Q is also = C*ΔV, where C is the capacitance and ΔV is the voltage drop as the current flows out.

=> CΔV = Q = It

Rearranging gives C = I*t/ΔV.

For a 2A power supply, 60Hz, full-wave, where you can tolerate a 12V sag in the filter capacitor voltage without the regulator dropping out of regulation,

C = 2 * 0.008/12 = 0.0053F = 1,333uF 

As I mentioned earlier that capacitor needs to be rated for at least a 200V. As such your capacitor will be bulky and will not be cheap.

However, that does not directly correct the fact that you are over-volting the motor.



Additionally you should consider adding a snubber circuit across the terminals of your motor. This will reduce EMI from the motor and arcing on the brushes caused by the commutator switching. The values of R&amp;C will depend again on the motor resistance and commutation frequency.