A small wooden block of density `400 kg//m^(3)` is released from rest inside a swimming pool filled with water of density `10^(3)kg//m^(3)`. The acceleration of the block just after it is released is __________ `m//s^(2) (g=10 m//s^(2))`

Explanation

To find the acceleration of the wooden block just after it is released in water, we can follow these steps:

### Step 1: Identify the Given Data
- Density of the wooden block, \( \rho_b = 400 \, \text{kg/m}^3 \)
- Density of water, \( \rho_w = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Determine the Forces Acting on the Block
When the block is submerged in water, two main forces act on it:
1. The weight of the block (\( W \)) acting downwards:
\[ W = mg = \rho_b V g \]
where \( V \) is the volume of the block.

2. The buoyant force (\( F_b \)) acting upwards, which is equal to the weight of the water displaced by the block:
\[ F_b = \rho_w V g \]

### Step 3: Calculate the Net Force Acting on the Block
The net force (\( F_{net} \)) acting on the block can be calculated as:
\[ F_{net} = F_b - W \]
Substituting the expressions for \( F_b \) and \( W \):
\[ F_{net} = \rho_w V g - \rho_b V g \]
Factoring out \( Vg \):
\[ F_{net} = Vg (\rho_w - \rho_b) \]

### Step 4: Apply Newton's Second Law
According to Newton's second law, the net force is also equal to the mass of the block times its acceleration (\( a \)):
\[ F_{net} = ma \]
The mass of the block (\( m \)) is given by:
\[ m = \rho_b V \]
Thus, we can write:
\[ Vg (\rho_w - \rho_b) = \rho_b V a \]

### Step 5: Simplify the Equation
Dividing both sides by \( V \) (assuming \( V \neq 0 \)):
\[ g (\rho_w - \rho_b) = \rho_b a \]
Now, solving for \( a \):
\[ a = \frac{g (\rho_w - \rho_b)}{\rho_b} \]

### Step 6: Substitute the Values
Substituting the known values:
\[ a = \frac{10 \, \text{m/s}^2 \times (1000 \, \text{kg/m}^3 - 400 \, \text{kg/m}^3)}{400 \, \text{kg/m}^3} \]
\[ a = \frac{10 \, \text{m/s}^2 \times 600 \, \text{kg/m}^3}{400 \, \text{kg/m}^3} \]
\[ a = \frac{6000 \, \text{m/s}^2}{400} \]
\[ a = 15 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the block just after it is released is \( 15 \, \text{m/s}^2 \).

---