A cube of wood 20cm on each side floats in water so that 7cm is above the surfac

ID: 1392610 • Letter: A

Question

A cube of wood 20cm on each side floats in water so that 7cm is above the surface of the water and 13cm is below.

a)What is the density of the wood? ( You must justify your answer)

b) How much force would have to be exerted on the wood to submerge it?

c) what volume of lead mass has to be placed on top of the block of wood so that it will just be totally submerged?

d)Suppose a 5kg mass of lead is now placed on top of the block of wood, how much of the wood would now be above the water?

Densities.in g/cm^3 Water=1 g/cm^3 Iron=7.8g/cm^3 Lead=11.3g/cm^3

side of cube = s = 20 cm = 0,2 m

a) Applying buoyancy law

dwood s3 = dwater * (13/20)*s3

dwood = 13*1/20 = 0.65 g/cm3

Density of wood = dwood = 0.65 g/cm3

b) Force to be exerted to submerge the wood = ( dwater - dwood ) s3g = ( 1000 - 650)*0.2^3 *9.81 = 27.468 N

c) Force = Vleaddlead g

Vlead = 27.468 / ( 11300 * 9.81) = 0.000247 m3 = 247.78 cm3

d) Let the height of wood inside water is h

5*9.81 + 650*(0.2^3)*9.81 = 1000*0.2^2*h*9.81

h = (5 + 650*0.2^3) / (1000*0.2^2) = 0.255 m = 25.5 cm

h > 20 cm

Amount of wood above water is zero.

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.