# physics problem help?

A completely submerged chunk of metal sinks in water with an

acceleration equal to 1/5 of g ( where g=9.8 m/s2 ). Find the

specific gravity of this metal. Ignore all drag effects in the fluid.

### 2 Answers

- billrussell42Lv 71 month ago
The weight in air is w₀ = mg

the weight in water is w₁ = mg/5

buoyancy force is the difference = mg – mg/5 = (4/5)mg

Specific gravity is the ratio of the density of the material to the density of water

the buoyancy force is the weight of the displaced water = m₂g = V₂ρg where ρ is the density of water, V₂ is the volume of the displaced water.

buoyancy force = (4/5)mg = V₂ρg

(4/5)m = V₂ρ

specific gravity = (m/V) / ρ

chunk's volume is the same as the volume of the displaced water

V = V₂

specific gravity = (m/V) / ρ

(4/5)m = Vρ

V = (4/5)m/ρ = 4m/5p

specific gravity = (m/(4m/5p)) / ρ

specific gravity = ((m/ρ) / (4m/5p))

specific gravity = (m/ρ) (5p/4m)

specific gravity = 5/4

density of fresh water at 20C = 0.998 g/cm³ = 0.998 kg/L

= 998 kg/m³ = 8.33 lb/gal = 62.1 lb/ft³

- Log in to reply to the answers

- Steve4PhysicsLv 71 month ago
You can shorten the working but here it is in detail:

Mass of metal = m. Mass of water displaced = M.

Volume of metal = volume of water displaced = V.

Density of water = ρ

Resultant force on metal F = weight – upthrust:

F = mg – U

Using 'F = ma' with a = g/5 gives:

mg – U = mg/5

U = (4/5)mg

Upthrust = weight of water displaced. So weight of water displaced is (4/5)mg:

Mg = (4/5)mg

M= (4/5)m

Since volume = mass/density, volume of water displaced is:

V = M/ρ

. .= (4/5)m/ρ (equation 1)

Call metal's specific gravity ‘s’, then its density = ρs. Since volume = mass/density:

V = m/(ρs) (equation 2)

From equations 1 and 2:

(4/5)m/ρ = m/(ρs)

4/5 = 1/s

s = 5/4

. .= 1.25

- Log in to reply to the answers