Question

A 500g piece of wood (specific gravity =.5) floats in water (density=1.0 g/cm^3). A. What fraction...

A 500g piece of wood (specific gravity =.5) floats in water (density=1.0 g/cm^3).


A. What fraction of its volume will be submerged?

B. A 500g iron weight (density = 7.9g/cm^3) is hung from the bottom of the floating piece of wood. What fraction of the wood is submerged now?

Homework Answers

Answer #1

here,

mass of wood , m1 = 500 g

volume of wooden block , v1 = m1 /p1 = 1000 cm^3

a)

equating the forces

buoyant force = weight

volume submerged * pw * g = m1 * g

volume submerged * pw * g = volume * p * g

volume submerged = volume * 0.5

volume submerged /total volume = 0.5

b)

mass of iron , m2 = 500 g

density of iron , p2 = 7.9 g/cm^3

equating the forces

buoyant force = weight

pw * g * ( volume submerged of wood block + volume of iron block) = (m1 + m2) * g

1 * ( volume submerged of wood block + m2/p2) = (500 + 500)

1 * ( volume submerged of wood block + 500/7.9) = (1000)

volume submerged of wooden block = 936.7 cm^3

the fraction of wood submerged now , f = submerged volume of wooden block /total volume

f = 0.94

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