Question

A wooden block of volume 5 * 10 ^ - 4 * m ^ 3 floats partially submerged in water, and a small steel object of mass m is placed on top of the block. When m = 0.25 kg, the system is in equilibrium and the top of the wooden block is at the level of the water. What is the density of the wood?

Answer 1

The density of the wooden block is equal to the density of water, which is approximately 1000 kg/m^3.

Step-by-step explanation:

To solve this problem, we need to consider the forces and equilibrium conditions involved in the system.

Given information:

Volume of the wooden block (V_block) = 5 * 10^-4 m^3

Mass of the steel object (m) = 0.25 kg

Let's assume:

Density of water (ρ_water) = 1000 kg/m^3 (standard value)

Density of the wooden block (ρ_wood) = ρ

Density of steel (ρ_steel) = 7850 kg/m^3 (standard value)

Volume of the wooden block submerged in water (V_submerged) = V_block_submerged

When the system is in equilibrium, the buoyant force on the wooden block must be equal to the weight of the steel object on top of it.

Buoyant Force (F_buoyant) = Weight of Steel Object (F_weight)

The buoyant force can be calculated using Archimedes' principle:

F_buoyant = ρ_water * g * V_submerged

The weight of the steel object can be calculated using its mass and the acceleration due to gravity:

F_weight = m * g

Setting these two forces equal to each other and solving for V_submerged:

ρ_water * g * V_submerged = m * g

V_submerged = m / ρ_water

Now, we know that the volume of the wooden block submerged in water is equal to the volume of the wooden block itself (V_block_submerged = V_block). Therefore:

V_block = m / ρ_water

We can also relate the density of the wooden block to its mass and volume:

ρ_wood = m / V_block

Substituting the expressions for V_block and V_submerged:

ρ_wood = ρ_water * (V_block / V_block) = ρ_water

So, the density of the wooden block is equal to the density of water, which is approximately 1000 kg/m^3.