There are forces are acting on the block: 1. Fs = upward force from scale (up = + Fs) 2. W = downward weight force (down = - W) 3. Fb = upward buoyancy force (up = + Fb)
Since the block is at rest, (net force = 0), the force equation is:
Fs + Fb - W = 0 ~(eq1)
Substituting Fs, Fb and W into ~(eq1): ==> {m (on scale) x g} + {ρ x V x g} - (m (block) x g} = 0
Dividing both sides of equation by g: m (on scale) + (ρ x V) - m (block) = 0 ~(eq2)
Given: V = volume of water displaced = 40% volume of the block = 0.40 x V (block)
Substituting 0.40 x V (block) into ~(eq2): m (on scale) + (ρ x 0.40 x V (block)) - m (block) = 0 ~(eq3) where, m (on scale) = 5.6 g ρ = density of water at 20ºC = 0.99821 g/cm^3 m (block) = 50 g
Substituting m (on scale), ρ and m (block) into ~(eq3): 5.6 g + (0.99821 g/cm^3 x 0.40 x V (block)) - 50 g = 0 ~(eq4)
Simplifying ~(eq4) and solving for V (block): ==> 5.6 g + 0.399284 g/cm^3 x V (block) - 50 g = 0 ==> 0.399284 g/cm^3 x V (block) - 44.4 g = 0 ==> 0.399284 g/cm^3 x V (block) = 44.4 g ==> V (block) = 44.4 g / (0.399284 g/cm^3) ==> V (block) = 111.199 cm^3
ρ (block) = density of block = m (block) / V (block) = 50 g / 111.199 cm^3 = 0.449644 g / cm^3
