Question

A spring is attched to the bottom of an empty swimming pool, with axis ot the spring oriented vertically. An 8.0 kg block of wood with density of d=850 kgm^-3 is fixed to the top of spring and compereeses it. Then the pool is filled with water, completley covering the block. The spring is now observed to be streched twice as much as it had been compressed. a) Determine the percentage of the block's total volume that is hollow. Ignor any air hollow space. (density of water d = 1000 kgm^-3) b) If the spring compressed 0.1 m find the spring constant, k.

Answer 1

First, volume displaced:
ρV = m ==> ρ/m = V
Where ρ is density, V is volume, m is mass

The block pushes down with a force equal to its weight, 8kgf. Since force =-kx for a spring, when the pool is filled, it lifts with a force of 16kgf against the spring. So, it must displace 92kg of water, because it must lift 16kgf for the spring force, and 16kgf for its own weight.

ρ∙V = m ==> ρ/m = V
total water V = ρ/m = 1000kg/m³ / 92kg = 10.868m³

wood V = ρ/m = 850kg/m³ / 8kg = 106.25m³

void V = ΔV = 106.25m³ - 10.868m³ = 85.38m³

% hollow = Vvoid/ Vtotal = 0.0452m³ / 0.075m³ = 60.3%

I use gravimetric units [kgf], which is not strictly allowed for SI; use mg = W instead of kgf.