I would not recommend that: it would result in enormous power wastage. The reason being, during the on part of the PWM, the circuitry (and cables, component leads etc) between the mains supply and the capacitor would be acting basically as a (very low value) resistor which would dissipate the heat. For example, the approximate average voltage across the MOSFET/cables/rectifier would be 200v. In other words, very inefficient. Lastly, There would have to be a current-limiting feature controlling the MOSFET, otherwise the surge in current would also blow fuses, because in essence it would just be one giant short circuit.
But yes, regarding the peak current through the MOSFET, the MOSFET will have graphs and other specs which will define separately, the peak current (for example over extremely short pulses) as well as a total power dissipation which if the MOSFET is pulsed on and off could essentially be doubled for your purposes. But calculations would have to take into account the effectiveness of what heatsink you have etc.
As the comments to your question say, you need to find out a suitable capacitor value. You can then see what the valley voltage of your ripple is, add in the voltage drop caused by the rectifier diodes, then arrive at the MINIMUM transformer secondary voltage. You will then select the next highest available transformer secondary voltage, re-do all of the calculations, and finally find out how much heat that you need to get rid of.
Note that you need to design for the minimum available AC Mains voltage available when choosing the minimum secondary voltage. You need to design for the MAXIMUM AC Mains voltage when calculating how much heat will be generated.
Engineering is all about trade-offs. If the available transformer secondary voltage is higher than you like, you can choose a smaller reservoir capacitor such that the worst-case ripple valley voltage is still above the regulator drop-out voltage. This extra ripple translates into less heat wasted in the regulator.
Do note that you must NOT allow the ripple valley voltage to drop below the regulator drop-out voltage or the resulting regulated DC supply will have AC Mains ripple present in the output. That can cause all manner of problems, especially in audio circuits.
Finally, pay attention to the ripple current rating of the reservoir capacitor when choosing a suitable part. Note that the peak ripple current is substantially higher than the DC current output of the supply - the capacitor has to charge in only a small portion of the incoming AC sine-wave waveform.
Best Answer
This is a tricky one. The max rating of the BLW77 is realistically 35 volts but lowering the voltage reasonably under 35V to (say) 32 volts by a resistor is going to dissipate a lot of power in the resistor at 8 amps current.
Power in resistor will be V.I = (35.7 - 32).8 = 29.6 watts.
Dropping it to 28 V is going to dissipate about 62 watts so I believe your best bet is to construct a buck regulator to drop anything between 30V to 40V down to a regulated 28 volts. Maybe one of these: -
You can set the output voltage to be 28V by adjusting the potential divider on the Vfb pin. For alternative try Linear technology and use their technical search engine for narrowing down results. TI also have a great range of buck regulators too.