Question

a person is floating on a raft, their total volume is 1.5 m³. the person and raft displace 0.224 m³ of water. what is the density of the person and raft combined? ? kg/m³ ρ_water = 1,000 kg/m³

Explanation:

Step1: Recall Archimedes' Principle (buoyant force equals weight of displaced fluid) and density formula.

Since the person and raft are floating, their weight equals the weight of displaced water. First, find the mass of displaced water using \( m =
ho V \), where \(
ho_{water} = 1000 \, \text{kg/m}^3 \) and \( V_{displaced} = 0.224 \, \text{m}^3 \). Then, the mass of the person and raft is equal to the mass of displaced water. Finally, use \(
ho = \frac{m}{V_{total}} \), where \( V_{total} = 1.5 \, \text{m}^3 \).

Step2: Calculate mass of displaced water.

\( m_{displaced} =
ho_{water} \times V_{displaced} = 1000 \, \text{kg/m}^3 \times 0.224 \, \text{m}^3 = 224 \, \text{kg} \).

Step3: Calculate density of person and raft.

The mass of the person and raft (\( m \)) is equal to \( m_{displaced} \) (since floating, \( F_b = mg = m_{displaced}g \), so \( m = m_{displaced} \)). Now, use \(
ho = \frac{m}{V_{total}} = \frac{224 \, \text{kg}}{1.5 \, \text{m}^3} \approx 149.33 \, \text{kg/m}^3 \)? Wait, no—wait, wait, no! Wait, the total volume of the person is 1.5 m³, but the person and raft's total volume? Wait, no, the problem says "their total volume is 1.5 m³" (person and raft? Wait, re-reading: "their total volume is 1.5 m³"—so person and raft combined volume is 1.5 m³? Wait, no, maybe misread. Wait, the problem: "A person is floating on a raft, their total volume is 1.5 m³"—maybe "their" (person and raft) total volume is 1.5 m³. And they displace 0.224 m³ of water. So when floating, mass of object = mass of displaced water. So mass of person+raft = mass of displaced water = \(
ho_{water} V_{displaced} = 1000 \times 0.224 = 224 \, \text{kg} \). Then density \(
ho = \frac{m}{V} = \frac{224 \, \text{kg}}{1.5 \, \text{m}^3} \approx 149.33 \, \text{kg/m}^3 \)? Wait, that seems low, but let's check again. Wait, maybe the "their total volume" is the person's volume? No, the problem says "A person is floating on a raft, their total volume is 1.5 m³"—"their" (person and raft) total volume. So yes, \( V = 1.5 \, \text{m}^3 \), displaced volume \( 0.224 \, \text{m}^3 \). So mass of displaced water: 1000*0.224=224 kg. Then density of person+raft: 224 kg / 1.5 m³ ≈ 149.33 kg/m³. Wait, but let's confirm the steps.

Wait, Archimedes' principle: buoyant force equals weight of displaced fluid. For floating objects, weight of object = weight of displaced fluid. So \( m_{object} g = m_{displaced} g \), so \( m_{object} = m_{displaced} \). Then density of object \(
ho_{object} = \frac{m_{object}}{V_{object}} = \frac{m_{displaced}}{V_{object}} \).

So \( m_{displaced} =
ho_{water} V_{displaced} = 1000 * 0.224 = 224 \, \text{kg} \).

\( V_{object} = 1.5 \, \text{m}^3 \).

Thus, \(
ho_{object} = \frac{224}{1.5} \approx 149.33 \, \text{kg/m}^3 \). Wait, but maybe I misread the volume. Wait, the problem says "their total volume is 1.5 m³"—"their" (person and raft) total volume. So yes, that's correct.

Wait, but let's check units again. \(
ho = \frac{m}{V} \), m in kg, V in m³, so kg/m³. Correct.

So steps:

1. Find mass of displaced water: \( m = 1000 * 0.224 = 224 \, \text{kg} \).

1. Density of person+raft: \(

ho = \frac{224}{1.5} \approx 149.33 \, \text{kg/m}^3 \). Wait, but maybe the total volume is the volume of the person, and the raft's volume is separate? No, the problem says "their total volume is 1.5 m³"—"their" (person and raft) so combined volume is 1.5 m³.

Yes, so the calculation is correct.

Answer:

\( \approx 149.33 \, \text{kg/m}^3 \) (or more precisely, \( \frac{224}{1.5} = \frac{2240}{15} = \frac{448}{3} \approx 149.33 \))