What is the magnitude of the net force on the cube? (use g=10 m/s^2)?

Ignoring drag, there are two forces on the cube--its own weight, and the buoyancy of the water.

The cube weighs 65 N (6.5 kg * 10 m/s^2). This is the downward force.

The buoyant force is a little harder to calculate, but not terribly so. Total volume is (.2m)^3, or .008 m^3. The mass of the displaced water is thus 1000 kg/m^3 * .008 m^3, or 8 kg, which makes the weight 80 N.

Archimedes' principle states that the buoyant force is equal to the weight of the displaced fluid. In other words, because the cube is displacing 80 Newtons' worth of water, there's a(n upward) buoyant force of 80 N on the cube.

So, taking up as positive, the force on the cube is (+80 N) + (-65 N), or 15 N.

Further notes: The distance that the cube is underwater doesn't come into play. It only matters that the cube is submerged (and not at some extreme depth that would affect the density of the water). A good free body diagram (FBD) is always helpful in solving problems involving more than one force acting on an object.

Hope this helps.

Archimedes's Principle says that the buoyant force is equal to the weight of the displaced water. Since the whole cube is submerged, the volume of displaced water is the volume of the cube.

V = (0.2)^3

V = .008 m^3

Mass of displaced water = 1000(.008) = 8kg

8(10) = 80 N of water

Buoyant force = 80 N (up)

Force of gravity is the mass multiplied by g

6.5(10) = 65 (down)

80 - 65 = 15 N