How much more energy would I spend if my bicycle was 2 kilos heavier? I have a bicycle weighing 8 kilos.  An 10 kilometer ride averaging 20 km/h requires z kWh's of energy. 
How much more energy would I spend if I added 2 kilos to my backpack?  
Or turning it around: If I used the same amount of energy, how much slower would I get?
UPDATE: I'm asking out of curiosity, not out of some homework assignment.  It's a long time since I've had those :) 
I have friends arguing that a 2 kg lighter bike (e.g. a racer instead of a cyclo cross) will give me serious advantages because of the decreased weight, even though I'm an amateur when it comes to bicycling. Me, on the other hand, can't really see why this would matter all that much until I get seriously more fit.  Anyway, I'm not familiar with the physics needed to calculate this.
If it helps, let's assume a 5 degree rise from 0-5 km in 10 km/h and a 5 degree descent from 5-10 km in 40 km/h.
 A: If you are doing this out of curiosity for a specific route:
The easiest way to get a rough estimate is to look at the elevation profile of your route and assume that when going downhill you are expending the same amount of energy regardless of the weight (which might not be accurate), and base all your energy changes when going uphill.  Your change in energy requirements on a specific uphill would be:
$$E_{change} = (m_{new}-m_{old})*g*h$$
If you know the average speed up the hill, your average power change during only that hill would be:
$$P_{change} = (m_{new}-m_{old})*g*V_{avg}$$
If you do this for all hills, and assume downhills and flats are the same power, you can average all your $P_{change}$ for all those sections and get your average $P_{change}$.
If your trip is mostly flat, my guess would be that your power savings would be almost unnoticeable, unless like somebody commented, you are just stopping and accelerating all the time while trying to keep the same average speed for both mass conditions.
If this is a homework question:
I think the way to do it is like yankeefan11's answer here using $E = 1/2*m*v^2$. The distance comes into play when you want to calculate the change in power.
A: We can use the fact that our energy will be given as
$$E = \frac12mv^2$$
So assuming that you can neglect air resistance and that you keep the same speed, and everything is kept simple, the additional energy you need is:
$$E_{new}-E_{old}$$
$$\frac 12m_{new}v^2-\frac 12 m_{old}v^2$$
$$\frac 12v^2(m_{new}-m_{old})$$
You know the difference in mass that you added, so you can just plug in your values.
