Electronic – Mosfets cooled with Heatsink and fan

To directly answer your first question: No, not even close. A little more copper around a surface mount part isn't going to get rid of 11W of heat. No way.

One answer may be to parallel multiple FETs. Not only does that cut the total dissipation by the number of parts, but the dissipation on each FET is reduced by the square of the number of parts. So if one FET dissipates 10W, then two parallel FETs would dissipate 5W total, and each FET would dissipate only 2.5W.

That's in theory. In practise they won't share the load exactly equally, so you have to design for a bit worse than that per FET. The good thing about paralleling FETs is that they have a positive temperature coefficient. The Rdson goes up with temperature. This helps them ballance somewhat and prevents runaway of a single part, like could be the case with bipolar transistors.

Ultimately you have to decide what you really want. Switching 40A is going to make some heat. One way or the other you're going to have to deal with that. You can tell us you're space contrained all you want, but ultimately the physics will dictate a certain amount of space, surface area, forced cooling, or whatever. It may not be possible to meet all the constraints. Not all combinations of small size, high current, and low cost are possible.

Answering your second question:

A switching MOSFET will have two types of losses; conduction and switching. Conduction loss is the usual \$I_D^2 \times R_{DS(on)}\$ loss. If you control the MOSFET so that it's on with 50% duty cycle, the conduction loss is 50% of the DC (always-on) loss.

Switching losses include the amount of energy needed to control the gate and losses in the device as it transitions from the on-state to the off-state. When you're turning on a MOSFET, there is an interval where \$I_D\$ starts to flow and the \$V_{DS}\$ voltage is still at its maximum. \$V_{DS}\$ falls as the MOSFET channel saturates. The power consumed during this time is called turn-on loss. Similarly, at turn-off, there's an interval where \$V_{DS}\$ rises before \$I_D\$ starts falling, which (not surprisingly) is called turn-off loss.

You must consider the turn-on and turn-off losses when you're talking about 100kHz operation. Most likely you will see less power than the DC condition, but you won't be saving 50%.

Answering your third question:

MOSFET \$R_{DS(on)}\$ has a positive temperature coefficient - the warmer it gets, the higher the \$R_{DS(on)}\$ gets. If you connect two MOSFETs in parallel with similar characteristics (i.e. the same part number from the same manufacturer), drive them identically, and don't have huge asymmetry in your PCB layout, the MOSFETs will indeed share current quite nicely. Always make sure each MOSFET has an independent resistor in series with each gate (never parallel gates without resistors) as gates tied directly together can weirdly interact with each other - even a few ohms is better than nothing.