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

Answer 1

a)

let the density of wood is p

Now,

0.20^3 * p * 9.8 = 0.20^2 * 0.13 * pwater * 9.8

0.20^3 * p * 9.8 = 0.20^2 * 0.13 * 1000 * 9.8

p = 650 Kg/m^3

the density of wood is 650 Kg/m^3

b)

force needed to submerge wood ,

W = weight of additional water displaced

W = 0.20^2 * 0.07 * 1000 * 9.8

W = 27.44 N

the force needed is 27.44 N

c)

let mass of lead needed is m

27.44 = m * 9.8

m = 2.8 Kg

volume of lead = 2.8/(11300)

volume of lead = 2.477 *10^-4 m^3

d)

as the mass of lead is greater than 2.8 Kg ,

the block of wood will sink