A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline. Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What is the mass of the wooden block? A.) 40 gB.) 20 gC.) 10 g

b) M=20g

Explanation:

For this exercise we must use the Archimedes principle that states that the thrust that a body receives is equal to the weight of the dislodged liquid.

B = ρ g V

Let's use balance healing for this case

Initial.

B - W = 0

The weight of the body can be related to its density

W = ρ V_body g

ρ_liq g (½ V_body) = m g

Final

Some masses were added

M = 20 g = 0.020 kg

B - W - W₂ = 0

ρ_liq g V_Body = m g + M g

Let's replace and write the system of equations

½ ρ_liq V_body = m

ρ V_body = m + M

We solve the equations

2 m = m + M

m = M

m = 20 g

The answer is b