A body suspended from a dynamometer stretches the spring with a force of 5N

A body suspended from a dynamometer stretches the spring with a force of 5N in air, and with a force of 3N in water. Determine the density of this body.

P = 5 N.

PB = 3 N.

ρw = 1000 kg / m3.

ρ -?

The body weight in the air is determined by the formula: P = m * g = ρ * V * g.

m = P / g.

The weight of the body Рв, which is immersed in water, is expressed by the formula: Рв = Р – Farch.

The buoyancy force of Archimedes Farch is determined by the formula: Farch = ρw * g * V. Where ρw is the density of the liquid in which the body is immersed, g is the acceleration of gravity, V is the volume of the immersed part of the body in the liquid.

Pw = P – ρw * g * V.

V = (Р – Рв) / ρв * g.

ρ = m / V = Р * ρв * g / g * (Р – Рв) = Р * ρв / (Р – Рв).

ρ = 5 N * 1000 kg / m3 / (5 N – 3 N) = 2500 kg / m3.

Answer: the body density is ρ = 2500 kg / m3.