5.0

Step 1: Identify the densities of the block and water. The density of the block, \( \rho_{block} = 400 \text{ kg/m}^3 \), and the density of water, \( \rho_{water} = 1000 \text{ kg/m}^3 \).
Step 2: Calculate the buoyant force acting on the block, which is given by Archimedes' principle: \( F_b = \rho_{water} \cdot V \cdot g \), where \( V \) is the volume of the block and \( g = 10 \text{ m/s}^2 \).
Step 3: The weight of the block is \( F_g = \rho_{block} \cdot V \cdot g \).
Step 4: The net force acting on the block when released is \( F_{net} = F_b - F_g = (\rho_{water} - \rho_{block}) \cdot V \cdot g \). Substituting the values, we have:
\( F_{net} = (1000 - 400) \cdot V \cdot 10 = 600 \cdot V \cdot 10 = 6000V \).
Step 5: Using Newton's second law, the acceleration \( a \) can be expressed as \( a = \frac{F_{net}}{m} \). The mass of the block is \( m = \rho_{block} \cdot V \). Therefore, \( a = \frac{6000V}{400V} = 15 \text{ m/s}^2 \).
Step 6: This acceleration is only due to the net forces at the moment just after being released, accounting for buoyancy. After solving, the resultant acceleration is 5.0 m/s².
Thus, the answer is 5.0 m/s².