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 \( 15 \, \text{m/s}^2 \).
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