Question

2. (10 pts) A piece of wood with a density Pwd06g is tied with a string to the bottom of a water-filled flask and is in equilibrium. The wood is completely immersed. The tension in the string is 0.89 N. (the density of water is pwater- 1.00 x 103 kg a. (4 pts) Find the volume of the wood. b. (3 pts) If the string breaks, the wood starts moving upward due to the buoyancy force. Neglecting any viscous drag of water, determine the acceleration of the wood block at the time the string breaks. c. (1pt) After reaching the top of the surface, the wood block floa ts on the surface, does the water level in the flask rise, drop or stay the same? (2 pts) Determine the fraction of wood's volume immersed while the wood surface of water.

Answer 1

here,

density of wood , p1 = 706 kg/m^3

the tension in the string , T = 0.89 N

density of water , p2 = 1000 kg/m^3

a)

let the volume of wood be V

tension in the string , T = buoyant force - m * g

T = p2 * V * g - p1 * V * g

0.89 = V * 9.81 * ( 1000 - 706)

solving for V

V = 3.1 * 10^-4 m^3

b)

the acceleration , a = net force /effective mass

a = (buoyant force - m * g )/m

a = (p2 * V * g - p1 * V * g) /(p1 * V)

a = 9.81 * ( 1000 - 706)/706

a = 4.1 m/s^2

c)

as some volume of wood is outside the water

the volume of displaced is less

and the level of water will DROP

D)

let the volume immeresed be V'

as it floats

buoyant force = weight

p2 * V' * g = p1 * V * g

V' /V = 706 /1000

V' /V = 0.706