A block of wood has a mass of `25g`. When a `5g` metal piece with a volume of `2cm^(3)` is attached to the bottom of the block, the wood barely floats in water. What is the volume `V` of the wood?

Explanation

To solve the problem, we need to find the volume \( V \) of the wooden block when a metal piece is attached to it, and the combination barely floats in water. Here's the step-by-step solution:

### Step 1: Understand the Problem
The wooden block has a mass of \( 25 \, \text{g} \) and the metal piece has a mass of \( 5 \, \text{g} \) with a volume of \( 2 \, \text{cm}^3 \). When the metal piece is attached, the total weight of the system must equal the buoyant force acting on it in water.

### Step 2: Write Down the Known Values
- Mass of the wooden block \( M = 25 \, \text{g} = 0.025 \, \text{kg} \)
- Mass of the metal piece \( m = 5 \, \text{g} = 0.005 \, \text{kg} \)
- Volume of the metal piece \( V_m = 2 \, \text{cm}^3 = 2 \times 10^{-6} \, \text{m}^3 \)
- Density of water \( \rho_w = 1000 \, \text{kg/m}^3 \)

### Step 3: Set Up the Equation for Buoyancy
The buoyant force \( F_b \) is equal to the weight of the water displaced, which can be expressed as:
\[ F_b = \rho_w \cdot g \cdot V_d \]
where \( V_d \) is the total volume displaced by the wooden block and the metal piece.

### Step 4: Total Weight of the System
The total weight \( W \) of the wooden block and the metal piece is:
\[ W = (M + m) \cdot g = (0.025 + 0.005) \cdot g = 0.030 \cdot g \]

### Step 5: Volume Displaced
The total volume displaced \( V_d \) is the sum of the volume of the wooden block and the volume of the metal piece:
\[ V_d = V + V_m = V + 2 \times 10^{-6} \, \text{m}^3 \]

### Step 6: Set the Buoyant Force Equal to the Weight
Since the block barely floats, the buoyant force equals the weight:
\[ \rho_w \cdot g \cdot (V + 2 \times 10^{-6}) = (M + m) \cdot g \]

### Step 7: Cancel \( g \) from Both Sides
We can cancel \( g \) from both sides:
\[ \rho_w \cdot (V + 2 \times 10^{-6}) = 0.030 \]

### Step 8: Solve for \( V \)
Now, substitute \( \rho_w = 1000 \, \text{kg/m}^3 \):
\[ 1000 \cdot (V + 2 \times 10^{-6}) = 0.030 \]
\[ V + 2 \times 10^{-6} = \frac{0.030}{1000} \]
\[ V + 2 \times 10^{-6} = 3 \times 10^{-5} \]
\[ V = 3 \times 10^{-5} - 2 \times 10^{-6} \]
\[ V = 3 \times 10^{-5} - 0.2 \times 10^{-5} = 2.8 \times 10^{-5} \, \text{m}^3 \]

### Step 9: Convert to cm³
To convert \( V \) to cm³:
\[ V = 2.8 \times 10^{-5} \, \text{m}^3 = 28 \, \text{cm}^3 \]

### Final Answer
The volume \( V \) of the wooden block is \( 28 \, \text{cm}^3 \).

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