A spring is attached to the bottom of an swimming pool, with the axis of the spring oriented vertically. An 8.00 kg block of wood `(rho=840 kh//m^(3))` is fixed to the top of the spring and compressed it. Then the pool is filled with water, completely covering the block. the spring is now observed to be stretched twice as much as it had been conpressed. Determine the percentage of the block's total volume that is hollow. Ignore any air in the hollow space.

Explanation

To solve the problem step by step, we need to analyze the forces acting on the block of wood when it is submerged in water and how the spring behaves under these conditions.

### Step 1: Understand the initial compression of the spring
When the block of wood is placed on the spring and compressed by a distance \( x \), the force exerted by the spring is given by Hooke's Law:
\[ F_{\text{spring}} = kx \]
where \( k \) is the spring constant. At this point, the spring force balances the weight of the block:
\[ kx = mg \]
where \( m = 8.00 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \).

### Step 2: Analyze the situation when the spring is stretched
When the pool is filled with water and the spring is stretched to a distance \( 2x \), the forces acting on the block include the buoyant force from the water and the weight of the block. The net force on the block can be expressed as:
\[ F = mg + F_{\text{spring}} = mg + k(2x) \]
The buoyant force \( F_{\text{buoyant}} \) acting on the block is equal to the weight of the water displaced by the block, which can be expressed as:
\[ F_{\text{buoyant}} = \rho_{\text{water}} V_{\text{block}} g \]
where \( \rho_{\text{water}} \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)) and \( V_{\text{block}} \) is the volume of the block.

### Step 3: Set up the equation for the stretched spring
At the stretched position, we can write:
\[ F_{\text{buoyant}} = mg + k(2x) \]
Substituting the expression for buoyant force:
\[ \rho_{\text{water}} V_{\text{block}} g = mg + k(2x) \]

### Step 4: Relate the two scenarios
From the first scenario, we know:
\[ kx = mg \]
Thus, we can substitute \( kx \) in the second equation:
\[ \rho_{\text{water}} V_{\text{block}} g = mg + 2mg = 3mg \]
This simplifies to:
\[ \rho_{\text{water}} V_{\text{block}} g = 3mg \]

### Step 5: Solve for the volume of the block
Rearranging gives:
\[ V_{\text{block}} = \frac{3mg}{\rho_{\text{water}} g} = \frac{3m}{\rho_{\text{water}}} \]
Substituting \( m = 8.00 \, \text{kg} \) and \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \):
\[ V_{\text{block}} = \frac{3 \times 8.00}{1000} = 0.024 \, \text{m}^3 \]

### Step 6: Calculate the volume of the wood
The volume of the wood can be calculated using its density:
\[ V_{\text{wood}} = \frac{m}{\rho_{\text{wood}}} = \frac{8.00}{840} \approx 0.00952 \, \text{m}^3 \]

### Step 7: Calculate the volume of the hollow space
The volume of the hollow space \( V_{\text{hollow}} \) is given by:
\[ V_{\text{hollow}} = V_{\text{block}} - V_{\text{wood}} = 0.024 - 0.00952 = 0.01448 \, \text{m}^3 \]

### Step 8: Calculate the percentage of the block's total volume that is hollow
The percentage of the block's total volume that is hollow can be calculated as:
\[ \text{Percentage of hollow volume} = \left( \frac{V_{\text{hollow}}}{V_{\text{block}}} \right) \times 100 = \left( \frac{0.01448}{0.024} \right) \times 100 \approx 60.33\% \]

### Final Answer
The percentage of the block's total volume that is hollow is approximately **60.33%**.