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To solve this problem, we need to analyze the forces acting on the system when the wood block is floating with the lead block attached. Here’s the breakdown:

### Given Data:
- Mass of wood block, \( m_w = 5.50 \, \text{kg} \)
- Density of wood, \( \rho_w = 678 \, \text{kg/m}^3 \)
- Density of lead, \( \rho_{\text{lead}} = 11340 \, \text{kg/m}^3 \)
- Density of water, \( \rho_{\text{water}} \approx 1000 \, \text{kg/m}^3 \) (assumed typical density for water)

### Step-by-Step Solution:

1. **Volume of the Wood Block**:
   The volume \( V_w \) of the wood block can be calculated using its mass and density:
   \[
   V_w = \frac{m_w}{\rho_w} = \frac{5.50 \, \text{kg}}{678 \, \text{kg/m}^3} \approx 0.00811 \, \text{m}^3
   \]

2. **Buoyant Force Acting on the System**:
   Since the upper surface of the wood is level with the water, the entire volume of the wood is submerged. The buoyant force \( F_b \) acting on the system is equal to the weight of the water displaced:
   \[
   F_b = \rho_{\text{water}} \cdot V_w \cdot g = 1000 \, \text{kg/m}^3 \times 0.00811 \, \text{m}^3 \times 9.8 \, \text{m/s}^2 \approx 79.48 \, \text{N}
   \]

3. **Total Weight of the System**:
   The total weight of the system (wood block + lead block) should be equal to the buoyant force for the system to be in equilibrium.

   Let the mass of the lead block be \( m_{\text{lead}} \). Then, the weight of the system is:
   \[
   W_{\text{total}} = (m_w + m_{\text{lead}}) \cdot g
   \]
   Setting \( W_{\text{total}} = F_b \):
   \[
   (5.50 + m_{\text{lead}}) \cdot 9.8 = 79.48
   \]
   Solving for \( m_{\text{lead}} \):
   \[
   m_{\text{lead}} = \frac{79.48}{9.8} - 5.50 \approx 2.6 \, \text{kg}
   \]

### Answer:
The mass of the lead block is approximately **2.6 kg**.

Would you like more details, or do you have questions on any specific part?

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### Additional Related Questions:
1. How would the solution change if the density of water were different, say in seawater?
2. What happens to the equilibrium if a portion of the wood block is above the water surface?
3. How would you calculate the volume of the lead block based on its mass?
4. How does this setup demonstrate Archimedes' Principle?
5. How would the solution change if the lead block were fully submerged in the water instead of hanging?

### Tip:
Always verify units at each step in physics problems to ensure dimensional consistency.

A lead block is suspended by means of a string from the underside of a 5.50 kg block of wood of density of 678 kg/m³. If the upper surface of the wood is just level with the water, what is the mass of the lead block? (The density of lead is 11340 kg/m³)

Learn how to calculate the mass of a lead block suspended from a wood block when the upper surface of the wood is level with water. This physics problem involves buoyancy, force equilibrium, and density calculations, and is suitable for students in Grades 11-12.

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