Question

A cube of wood with a density of 0.780 g/cm 3 is 10.0 cm on each side. When the cube is placed in water, what buoyant force acts on the wood

Answer 1

Answer:

The buoyant force on the wood = 7.652 N

Explanation:

According to the principle of flotation, a body floats when the upthrust exerted upon it by the fluid n which it floats equals the weight of the body.

W = U ............... Equation.

Where W = weight of the wood, U = Upthrust or buoyant force.

Recall That,

Density = mass/volume

Mass = Density×volume

m = D×V........................ Equation 2

Where m = mass of the wood, V = Volume of the wood, D = Density of the wood.

But

Volume of a cube = a³

V = a³ where a = length of the cube.

V = 10³ = 1000 cm³.

Given: V = 1000 cm³ D = 0.780 g/cm³

Substituting these values into equation 2,

m = 1000(0.780)

m = 780 g

m = 0.78 kg.

But W = mg

Where m = 0.78 kg, g = 9.81 m/s²

W = 0.78(9.81)

W = 7.652 N.

Since W = U = 7.652 N.

U = 7.652 N

Therefore the buoyant force on the wood =7.652 N