How fast will a styrofoam cube with lift force of 228kg rise , when submerged 10 meters underwater

Question

the dimensions of the cube are L 75cm H 180 cm W 15 cm. the area of the side of the cube that will be pointing upward is 1125 cm^2. how much time will be needed fot the cube to surface to the top. what will be the result if the cube was submerged at 20 meters underewater. thank you

Answer 1

Answering this question depends on how quickly the cube reaches its terminal velocity on the way up and how linearly it behaves on it's accent.  However you don't really have a cube do you? :-) Nevertheless:

1. We need to know  the buoyant force Fb the cube experiences while submerged in water

Fb=gV[p(water) - p (styrofoam)] or

Fb=gV[p1 - p2]

2. The terminal velocity Vt  [1] is given as

Vt= sqrt( 2 m a/p1 A Cd)

where

Vt = terminal velocity,

m = mass of the falling object,

a = acceleration due to the buoyant force,

Cd = drag coefficient,

p1 = density of the water

A = projected (or effective) area of the object.

in this case acceleration is

a=Fb/m

Since the object is of a bit irregular shape its terminal velocity is a bit difficult to establish. Considering that the styrofoam object reaches its terminal velocity fast the  motion governing equation however is

S= Vt t

thus the time is

t= S/Vt

To compute Vt= sqrt( 2 m a/p1 A Cd)

Let

Cd= 1.2

A= 1125 cm^2

m= p2 V= 60kg/m^3 x (75x 180x 15 )/(100^3)

m= 12 kg

a= Fg/m= g V ( p1 - p2)/ m= 9.81 x 0.2025 x (1000 - 60) / 12

a= 156 m/s^2

A= 1125 cm^2 = 0.1125 m^2

Now

Vt= sqrt( 2 x12 x  156/[1000  0.1125  1.2])

Vt= 5.3 m/s

Now time that will take the object to surface

10 m

t(10)= 10/ 5.3 =1.9 sec

20m

t(20)=20/5.3=3.8 sec

That does it!