Calculate the following

(1) 30

(2) 0.48

(3) 38.7

Explanation

1. What is the density of the wood?

1.Recall the facts given

• Density of water = 62.4 lb/ft³
• A cubic foot of pine wood weighs about 30 lb/ft³

That means the density of pine wood is \(30 \,\text{lb/ft}^3\).

2. Volume of the wood block

The given block has

$V = 0.48 \,\text{ft}^3$

3. Weight of the wood

$W = \text{density of wood} \times V$

$W = 30 \,\frac{\text{lb}}{\text{ft}^3} \times 0.48 \,\text{ft}^3 = 14.4 \,\text{lb}$

4. Buoyant force on the block

$F_b = \text{density of water} \times V$

$F_b = 62.4 \,\frac{\text{lb}}{\text{ft}^3} \times 0.48 \,\text{ft}^3 = 29.95 \,\text{lb} \approx 30 \,\text{lb}$

5. Effective density of the wood

The density is just its weight per unit volume:

$\rho_\text{wood} = \frac{W}{V} = \frac{14.4}{0.48} = 30 \,\frac{\text{lb}}{\text{ft}^3}$

✅ Answer: The density of the pine wood is 30 lb/ft³, consistent with the given fact.

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2. What is the density divided by the density of water?

Step 1. Density of wood

$\rho_\text{wood} = 30 \,\text{lb/ft}^3$

Step 2. Density of water

$\rho_\text{water} = 62.4 \,\text{lb/ft}^3$

Step 3. Ratio (specific gravity)

 

$\frac{\rho_\text{wood}}{\rho_\text{water}} = \frac{30}{62.4} \approx 0.48$

✅ Answer = 0.48

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3. Suppose a piece of oak wood flats with 62 percent of its volume underwater.  What is the density of the wood?

Step 1. Principle of flotation

A floating body displaces a volume of water equal to its own weight.

$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{wood}}}{\rho_{\text{water}}}$

Step 2. Given values

• Fraction submerged = 62% = 0.62
• Density of water = 62.4 lb/ft³

Step 3. Compute density of oak

$\rho_{\text{wood}} = 0.62 \times \rho_{\text{water}}$

$\rho_{\text{wood}} = 0.62 \times 62.4 = 38.7 \,\text{lb/ft}^3$

✅ Answer: The density of the oak wood is about 38.7 lb/ft³.