Question

A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface?

Answer 1

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

What mass of lead should be placed on the cube?

Given: Side of the cube (a) = 20cm

The density of the cube (ρc) = $650 kg/m^3$

a) Applying the force balance, the buoyant force must be equal to the weight of the cube

ρcgV = ρg × (Ax)

Substituting the values in the above equation, we get

$(650*(0.2 m)^3)=1000*(0.2 m)^2*x$

x = 0.13

where x is the height of the cube in the water

$A = a^2$ is the area of the cross-section

ρ is the density of the water

V is the volume of the cube

Now, the height above the surface of the water would be

h = a − x

Substituting the values, then we get

h = 0.2 − 0.13

h = 0.07 m

b) The mass added is "m" so the complete cube is submerged in the water, therefore

ρcgV + mg = ρg × (V)

$(650*(0.2 m)^3)+m=1000*(0.2 m)^3$

m = 2.8 kg

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

To learn more about buoyant force refer to:

brainly.com/question/11884584

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