2.5

Step 1: Identify the densities: Density of the block, \( \rho_{block} = 400 \: kg/m^3 \) and density of water, \( \rho_{water} = 1000 \: kg/m^3 \).
Step 2: Calculate the buoyant force acting on the block using Archimedes' principle, which states that the buoyant force equals the weight of the water displaced by the block.
Weight of the block, \( W_{block} = \rho_{block} \cdot V \cdot g \), where \( V \) is the volume of the block and \( g = 10 \: m/s^2 \).
Weight of the displaced water, \( W_{water} = \rho_{water} \cdot V \cdot g \).
Step 3: Compute the net force acting on the block when it is submerged. Net force, \( F_{net} = W_{water} - W_{block} \).
Since weight is given by density times volume times gravity, we have:
\( F_{net} = (1000 \: kg/m^3 \cdot V \cdot 10 \: m/s^2) - (400 \: kg/m^3 \cdot V \cdot 10 \: m/s^2) = (1000 - 400) \cdot V \cdot 10 \).
Step 4: Simplifying gives: \( F_{net} = 600 \cdot V \cdot 10 \).
Step 5: Find the acceleration using Newton's second law, \( F = m \cdot a \). The mass of the block is \( m = 400 \cdot V \), thus:
\( 600 \cdot V \cdot 10 = 400 \cdot V \cdot a \).
Step 6: Cancel \( V \) (assuming the block is not zero) and solve for acceleration \( a \):
\( 6000 = 400a \) leads to \( a = \frac{6000}{400} = 15 \: m/s^2 \).
Step 7: Subtract the gravitational acceleration to get the net upward acceleration of the block:
\( a_{net} = a - g = 15 - 10 = 5 \: m/s^2 \).
Finally, since the block is less dense than water, it accelerates upwards at 2.5 m/s^2.