Question

What mass (in kg) of iron must be placed on top of 3.47 kg the wood block so that the wood block is just barely submerged (the top of the block of wood is level with the surface of the water and the iron on top is in the air). Draw a picture of the situation before you attempt to solve the problem. The density of the wood is 620 kg/m3 and the density of the water is 1000 kg/m3.

Answer 1

M=2.12 Kg

Step-by-step explanation:

Given that

m= 3.47 kg

`Density\ of\ wood = 620\ kg/m^3`

`Density\ of\ water = 1000\ kg/m^3`

We know that

mass = volume x density

3.47 = 620 x V

`V=0.0055\ m^3`

Lets M mass put on wood block to submerged

So

Mg + mg = Fb

Fb = Bouncy force

`F_b= \rho_w* g* V`

Fb= 1000 x 0.0055 x 10 (take
`g=10\ m/s^2`)

Fb=55.96 N

Mg + mg = Fb

M x 10 + 3.47 x 10 =55.96

M=2.12 Kg