To find the density of the wood, we can use Archimedes' principle, which states that the buoyant force on the block is equal to the weight of the displaced water. In this case, the block of wood is floating, meaning the weight of the wood is balanced by the buoyant force.

Let’s go step by step:

Variables and Formula:

• Let $V_{\text{total}}$ be the total volume of the block of wood.
• $V_{\text{sub}}$ is the volume submerged under the water.
• The density of water is $\rho_{\text{water}} = 1000 \, \text{kg/m}^3$.
• The density of wood is $\rho_{\text{wood}}$.

Since 3/5 of the block's volume is above the water, the remaining 2/5 is submerged. So, we have: $V_{\text{sub}} = \frac{2}{5} V_{\text{total}}$

For the block to float, the weight of the wood must equal the weight of the water displaced. This is expressed by: $\text{Weight of wood} = \text{Weight of displaced water}$

The weight of the wood is: $\text{Weight of wood} = \rho_{\text{wood}} \cdot V_{\text{total}} \cdot g$ The weight of the displaced water is: $\text{Weight of displaced water} = \rho_{\text{water}} \cdot V_{\text{sub}} \cdot g$

Since these two weights are equal, we can write: $\rho_{\text{wood}} \cdot V_{\text{total}} \cdot g = \rho_{\text{water}} \cdot \left( \frac{2}{5} V_{\text{total}} \right) \cdot g$

Now, canceling out $V_{\text{total}}$ and $g$ from both sides: $\rho_{\text{wood}} = \rho_{\text{water}} \cdot \frac{2}{5}$

Substituting $\rho_{\text{water}} = 1000 \, \text{kg/m}^3$: $\rho_{\text{wood}} = 1000 \cdot \frac{2}{5} = 400 \, \text{kg/m}^3$

Conclusion:

The density of the wood is 400 kg/m³.

Would you like further details or have any questions?

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Here are 5 related questions for deeper understanding:

1. How would the density change if the block floated with 4/5 of its volume above water?
2. How does Archimedes' principle apply to objects that sink rather than float?
3. Can you explain how the gravitational acceleration $g$ cancels out in the equation?
4. What would happen to the density if the wood were floating in a different liquid (e.g., oil)?
5. How do we derive the formula for the weight of displaced water?

Tip: When dealing with buoyancy problems, always remember that the ratio of the submerged volume to the total volume gives important clues about the relative densities of the object and the fluid.