To solve the problem step by step, we will follow the principles of buoyancy and the relationship between weight, density, and volume.
### Step 1: Calculate the weight of the wooden block
The weight (W) of the wooden block can be calculated using the formula:
\[
W = m \cdot g
\]
where:
- \( m \) is the mass of the wood,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
First, we need to find the mass of the wooden block using its volume and density:
\[
m = V \cdot \rho
\]
Given:
- Volume of wood, \( V = 0.6 \, \text{m}^3 \)
- Density of wood, \( \rho = 800 \, \text{kg/m}^3 \)
Calculating the mass:
\[
m = 0.6 \, \text{m}^3 \cdot 800 \, \text{kg/m}^3 = 480 \, \text{kg}
\]
Now, calculating the weight:
\[
W = 480 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 4704 \, \text{N}
\]
### Step 2: Calculate the volume of water displaced
According to Archimedes' principle, the weight of the water displaced by the wooden block is equal to the weight of the block when it is floating.
The volume of water displaced (V_d) can be calculated using:
\[
V_d = \frac{W}{\rho_w \cdot g}
\]
where:
- \( \rho_w \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)).
Calculating the volume of water displaced:
\[
V_d = \frac{4704 \, \text{N}}{1000 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2} = 0.48 \, \text{m}^3
\]
### Step 3: Calculate the volume of the wooden block exposed
The volume of the wooden block that is exposed above the water surface can be calculated as:
\[
V_{\text{exposed}} = V - V_d
\]
where:
- \( V = 0.6 \, \text{m}^3 \) (total volume of the wooden block),
- \( V_d = 0.48 \, \text{m}^3 \) (volume of water displaced).
Calculating the exposed volume:
\[
V_{\text{exposed}} = 0.6 \, \text{m}^3 - 0.48 \, \text{m}^3 = 0.12 \, \text{m}^3
\]
### Step 4: Calculate the additional force required to submerge the block
To find the additional force (F) required to completely submerge the block, we can use the formula:
\[
F = V \cdot \rho_w \cdot g - W
\]
where:
- \( V = 0.6 \, \text{m}^3 \) (volume of the block),
- \( \rho_w = 1000 \, \text{kg/m}^3 \) (density of water),
- \( W = 4704 \, \text{N} \) (weight of the block).
Calculating the additional force:
\[
F = (0.6 \, \text{m}^3 \cdot 1000 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2) - 4704 \, \text{N}
\]
\[
F = (5880 \, \text{N}) - 4704 \, \text{N} = 1176 \, \text{N}
\]
### Final Answer
- Volume exposed: \( 0.12 \, \text{m}^3 \)
- Additional force required to submerge: \( 1176 \, \text{N} \)
