Question

A block of wood has a mass of 3.57 kg and a density of 588
kg/m^{3}. It is to be loaded with lead so that it will
float in water with 0.82 of its volume immersed. The density of
lead is 1.13 ? 10^{4} kg/m^{3}.

(a) What mass of lead is needed if the lead is on top of the
wood?

(b) What mass of lead is needed if the lead is attached below the
wood?

Answer #1

V = volume = 3.57/588 = 6.07*10^-3 m^3

v1 =volume inside water = 0.82*V = 4.97*10^-3 m^3

P = density of water =1000 kg/M^3

a) m1 = V1*P - 3.57 = V1*1000 -3.57 = 4.97 -3.57 = 1.4 kg

b) Let V' = volume of lead below the wood

m' = mass of lead having volume V'

m' = 1.13*10^{4} *V'

then 3.57 +m' = (V' + V1)*1000

3.57 + 1.13*10^{4} *V' = V' *1000 + 4.97

V' = 1.4/(10300) =1.359*10^{-4} m^3

m' = mass of lead= 1.13*10^{4} *V' =
1.53 kg

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