You are confused as to how transistors work. You may want to ground the emitter if this is a low side switch, but you certainly don't want to ground the base. The base is the input signal to the transistor.
Another problem is that the base will likely require more current than a microcontroller digital output can source. You want to switch 5 A, which will require a reasonable "power transistor". Even being generous and saying you get one with a gain of 50 at 5 A, that still requires 100 mA base current. You could use another transistor in emitter follower configuration driving the power transistor, or perhaps use a FET. Knowing what the voltage will be on the switch when open is useful in deciding the best strategy.
Now that you say you only need to switch 12 V, a FET will be a better fit. At that low voltage, you can find FETs that can turn on well with only 5 V on the gate. That means they can be driven directly from a 5 V digital logic output. In fact, these are sometimes called "logic level" FETs. 5 A is pretty high, but you may be able to find a FET that can do that with 5 V gate drive (I haven't looked). If not, you can definitely get this by paralleling two FETs. Unlike bipolar transistors, MOSFETs can be paralleled since their D-S resistance goes up with temperature. I use the IRLML2502 a lot for these kind of low voltage but decent current switching applications. A single IRLML2502 can't handle the current, but two in parallel will be fine. However, I would first look to see if you can find one that can do the whole job.
I haven't looked up the TIP3055, but just from it's name I'm guessing its a 2N3055 variant. That is a high current power transistor, but also has very low gain. You may need as much as 500 mA base current to get 5 A collector current. Driving that from a digital output will take additional active parts. The FET solution will be much simpler and will dissipate less power. You can probably get away without a heat sink with the right FET or FETs.