A cube of wood having an edge dimension of 20.8 cm and a density of 648 kg/m 3 f

Question

A cube of wood having an edge dimension of 20.8 cm and a density of 648 kg/m3 floats on water.

help fill in the blank please

Part 4 of 4 - Analyze The buoyant force supports the weight of both blocks, so we have B=Fg + Mg, where M is the mass of the lead. The buoyant force of the displaced volume of water is equal to the buoyant force of the submerged wood plus the weight of the lead. Using the expression for buoyant force pVg we have the following equation. Solving for the mass of the lead, we have 1000 Your response differs from the correct answer by more than 100%. kg/m3)( 648 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully, m ) = 2.72 Your response differs from the correct answer by more than 10%. Double check your calculations. kg. ] = Submit Skip (you cannot come back)

Explanation / Answer

given cube of wood of edge dimension a = 0.208 m
density of wood, rho = 648 kg/m^3

so there is another block of lead of mass M on top of this block
So the total buoyant force B = Fg + Mg
now if the wooden block is completely submerged in water
B = rho(water)*a^3*g
Fg = rho(wood)*a^3*g

hence
rho(water)a^3*g = rho(wood)a^3*g + Mg
1000*0.208^3 = 648*0.208^3 + M

M = (1000 - 648)*0.208^3 = 352*0.008998912 = 3.167617024 kg

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