MOSFET Heat dissipation question
Static dissipation with 4.5V drive will be something like 1.3W* maximum at 5A. They don't bother specifying the thermal resistance junction-to-air, but it's typically something like 65°C/W for a TO-220, so in a 50°C enclosure (easy on a warm day) it could be running at 135°C Tj, which is well above the absolute maximum. Typically, if you are lucky, not conservative, the MOSFET will be running at 'only' 116°C, which is just within the max rating. But we've not started to switch yet. Worst case will be when the LED strip is almost at full brightness. 

(By the way, it will be quite a bit better if you are using 10V drive- more like 0.8W maximum, which would be acceptable, at least in some undemanding applications (not if you need really high reliability or have to deal with hot ambients). These are designs based on worst-case specs, if you roll the dice most of the time you'll do better.) 

However you have to add the switching losses- they will depend on how hard you drive the gate. If you are using a lowish PWM frequency such as 120Hz and a powerful 10V gate driver you will minimize the losses, but if you trying to drive it directly with a microcontroller port pin, I think you'll have excessive losses. It's possible to calculate the losses based on the gate charge, PWM frequency and available current from your driver. If you are designing a real product you should do so. There is a small amount of capacitance from the gate to the drain, which is magnified by the Miller effect as the MOSFET switches. Your driver has to supply that current plus charge the gate capacitor.

* Conduction power dissipation is simply \$Pd=I^2Rds(on)\$ but the maximum value of Rds(on) will be higher by as much as 50% if the junction is hot.