Question

A man of mass m= 63.0 kg and having a density of p= 1060 kg/‘^3 (while holding his breath) is completely submerged in water. (A) Write newtons second law for this situation in terms of the man’s mass M, the density of water PW, his volume V, and G. Neglect any vicious drag of the water. May= (B) substitute M= PV into newtons second law and solve for the acceleration a, canceling common factors.ay=

(C). Calculate the numeric value of the man’s acceleration, assume upward as being the positive Y direction. (In m/s^2) (D) how long does it take the man to sink 7.40 m to the bottom of the lake? (In seconds)

Answer 1

The total force acting on the man is
=> Ft = Fb – m *g = ρw • V • g – m • g = ( ρw • V – m ) • g = m • a
here Fb = buoyant force

m = ρ • V
=> [ ρw • V – ρ • V ] • g = [ ρw – ρ ] * V * g = ( ρ *V ) * a

=>a = [ ρw – ρ ] * g / ρ = [ ( ρw / ρ ) - 1 ] * g
= [ ( 1000 / 1060 ) - 1 ] ( 9.8 ) = -0.55 in m / s ²
negative, Therefore the man sinks to the bottom
=> y = ½ a t ²

=>t = √ [ 2 y / a ] = √ [ 2 ( 7.40 ) / ( 0.55 ) ]

=5.19
Hence the man sinks to the bottom in 5.19 s