A block of wood, 10cm by 20 cm by 4 cm thick is floated on a pool of water, submerging it to a depth of 1 cm.  A 50 gram mass is placed on top of the block, increasing the depth submerged to 1.25.  Another 50 gram mass is placed on top, increasing the depth to 1.5.

What would the mass of the block of wood be?

To determine the mass of the block of wood, we can apply Archimedes' principle, which states that the buoyant force exerted on an object is equal to the weight of the fluid displaced by the object.

### Variables:
- Dimensions of the block: \(10 \, \text{cm} \times 20 \, \text{cm} \times 4 \, \text{cm}\)
- Density of water: \( \rho_{\text{water}} = 1 \, \text{g/cm}^3 \)
- First submersion depth: \( 1 \, \text{cm} \)
- Second submersion depth: \( 1.25 \, \text{cm} \) (after adding a 50 g mass)
- Third submersion depth: \( 1.5 \, \text{cm} \) (after adding another 50 g mass)

### Step 1: Buoyant Force at 1 cm Submersion
At 1 cm submersion depth, the block floats in equilibrium, so the buoyant force is equal to the weight of the block.

The volume of water displaced when submerged to 1 cm is:
\[
V_{\text{displaced}} = 10 \, \text{cm} \times 20 \, \text{cm} \times 1 \, \text{cm} = 200 \, \text{cm}^3
\]

Since the density of water is \(1 \, \text{g/cm}^3\), the mass of the displaced water is 200 g, which corresponds to the buoyant force:
\[
\text{Buoyant Force} = 200 \, \text{g} = \text{Mass of the block}
\]

So, the mass of the block is 200 g.

### Step 2: Adding the First 50 g Mass
After adding a 50 g mass, the depth of submersion increases to 1.25 cm. The new volume of water displaced is:
\[
V_{\text{displaced}} = 10 \, \text{cm} \times 20 \, \text{cm} \times 1.25 \, \text{cm} = 250 \, \text{cm}^3
\]

The corresponding buoyant force is:
\[
\text{Buoyant Force} = 250 \, \text{g}
\]

This buoyant force supports the weight of the block (200 g) plus the added 50 g mass, which is consistent with:
\[
200 \, \text{g} + 50 \, \text{g} = 250 \, \text{g}
\]

### Step 3: Adding the Second 50 g Mass
After adding another 50 g mass, the depth of submersion increases to 1.5 cm. The new volume of water displaced is:
\[
V_{\text{displaced}} = 10 \, \text{cm} \times 20 \, \text{cm} \times 1.5 \, \text{cm} = 300 \, \text{cm}^3
\]

The corresponding buoyant force is:
\[
\text{Buoyant Force} = 300 \, \text{g}
\]

This buoyant force supports the weight of the block (200 g) plus the two 50 g masses:
\[
200 \, \text{g} + 50 \, \text{g} + 50 \, \text{g} = 300 \, \text{g}
\]

### Conclusion:
The mass of the block of wood is \( \boxed{200 \, \text{g}} \).

Would you like more details on any part of this explanation, or do you have any additional questions?

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**Related Questions:**
1. What is the density of the block of wood?
2. How would the mass change if the block were submerged to 2 cm?
3. What if the block was made of a denser material like metal?
4. How does Archimedes' principle apply to floating objects of irregular shapes?
5. How would this problem change in saltwater, which has a higher density than freshwater?

**Tip:** Always ensure the units for volume and density are consistent when calculating buoyant forces.

A block of wood, 10cm by 20 cm by 4 cm thick is floated on a pool of water, submerging it to a depth of 1 cm. A 50 gram mass is placed on top of the block, increasing the depth submerged to 1.25 cm. Another 50 gram mass is placed on top, increasing the depth to 1.5 cm. What would the mass of the block of wood be?

Learn how to calculate the mass of a wooden block using Archimedes' principle. This problem involves buoyancy, volume displacement, and density of water. Suitable for students in Grades 8-10, this step-by-step solution explains key concepts and provides the necessary formulas.

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