A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (densityh=1.42 x 103 kg/m³) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure-not to scale). What is the magnitude of the buoyant force on the cube from the honey?
A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (densityh=1.42 x 103 kg/m³) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure-not to scale). What is the magnitude of the buoyant force on the cube from the honey?
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![**Question:**
A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (density \(\rho_h = 1.42 \times 10^3 \, \text{kg/m}^3\)) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure—not to scale). What is the magnitude of the buoyant force on the cube from the honey?
**Answer Choices:**
A) 13,900 N
B) 0 N
C) 78 N
D) 111 N
**Figure Explanation:**
The figure illustrates a cube submerged in honey, with the top surface of the cube situated 2.00 meters below the surface of the honey. The side of the cube is 0.200 meters long. A stick appears to suspend the cube within the honey.
**Solution:**
The magnitude of the buoyant force (\(F_b\)) on the cube can be determined using Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced fluid.
First, calculate the volume (\(V\)) of the cube:
\[ V = \text{side}^3 = (0.200 \, \text{m})^3 = 0.008 \, \text{m}^3 \]
Next, calculate the weight of the honey displaced by the cube. The density of the honey (\(\rho_h\)) is given as:
\[ \rho_h = 1.42 \times 10^3 \, \text{kg/m}^3 \]
The weight of the honey displaced is the product of the volume of the cube, the density of the honey, and the acceleration due to gravity (\(g\)), which is approximately \(9.8 \, \text{m/s}^2\):
\[ F_b = \rho_h \times V \times g \]
\[ F_b = (1.42 \times 10^3 \, \text{kg/m}^3) \times (0.008 \, \text{m}^3) \times (9.8 \, \text{m/s}^2) \]
\[ F_b = 111.104 \, \text{N} \]
Therefore](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fc0f41c-1d8b-41a4-aee9-3a82e03a5afd%2Fbc2e2963-ddf4-4cf6-963a-44344aaa4076%2F5erjx7m.png&w=3840&q=75)
Transcribed Image Text:**Question:**
A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (density \(\rho_h = 1.42 \times 10^3 \, \text{kg/m}^3\)) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure—not to scale). What is the magnitude of the buoyant force on the cube from the honey?
**Answer Choices:**
A) 13,900 N
B) 0 N
C) 78 N
D) 111 N
**Figure Explanation:**
The figure illustrates a cube submerged in honey, with the top surface of the cube situated 2.00 meters below the surface of the honey. The side of the cube is 0.200 meters long. A stick appears to suspend the cube within the honey.
**Solution:**
The magnitude of the buoyant force (\(F_b\)) on the cube can be determined using Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced fluid.
First, calculate the volume (\(V\)) of the cube:
\[ V = \text{side}^3 = (0.200 \, \text{m})^3 = 0.008 \, \text{m}^3 \]
Next, calculate the weight of the honey displaced by the cube. The density of the honey (\(\rho_h\)) is given as:
\[ \rho_h = 1.42 \times 10^3 \, \text{kg/m}^3 \]
The weight of the honey displaced is the product of the volume of the cube, the density of the honey, and the acceleration due to gravity (\(g\)), which is approximately \(9.8 \, \text{m/s}^2\):
\[ F_b = \rho_h \times V \times g \]
\[ F_b = (1.42 \times 10^3 \, \text{kg/m}^3) \times (0.008 \, \text{m}^3) \times (9.8 \, \text{m/s}^2) \]
\[ F_b = 111.104 \, \text{N} \]
Therefore
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