Question

A man of mass m=74.0 kg is holding his breath and floating, at rest, on the surface of a lake. While holding his breath, his density is p=994.0 kg.m^3. He exhales, which increases his density to 1030.0 kg/m^3 and so he begins to sink. The density of water is 1000.0 kg/m^3.

a) when the man is floating at the surface of lake, what percent of his body's volume is above water?

b) once the man exhales, what is the buoyancy force on the man?

c) After the man has been fully submerged in water, what will his acceleration be?

d) how long will it take the man to sink to the bottom of a 10.0 m lake?

Answer 1

mass , m = 74 kg

density of man when holding his breath , p₁ = 994 kg/m³

density of man when exhales , p₂ = 1030 kg/m³

density of water , p_w = 1000 kg/m³

a)

let the volume of body be V

let the volume of body above water be V'

equating the forces

buoyant force = weight

(V - V') * p_w * g = V * p₁ * g

( V - V') * 1000 = V * 994

V' = 0.006 V

V' /V = 0.006

the % of volume above water , % = V' /V * 100 = 0.6 %

b)

the buoyant force on man , F_b = p_w * V * g

Fb = pw * ( m/p₂) * g

Fb = 1000 * ( 74 /1030) * 9.81 = 704.8 N

c)

the accelration , a = (m * g - F_b)/m

a = ( 74 * 9.81 - 704.8) /74

a = 0.29 m/s² downwards

d)

let the time taken be t

h = 0 + 0.5 * a * t²

10 = 0 + 0.5 * 0.29 * t²

solving for t

t = 8.37 s