Part a) From the force balance, you know that ρ_{s}V_{s}g = ρ_{w}(1/3)V_{s}g , so the solid (s) density is 1/3 that of water.

From the definition of density, you know that ρ_{s} = m_{s}/V_{s} or that V_{s} = 3kg / ρ_{s} in m^{3}.

From the equation of volume of a cylinder V= πr^{2}h = (1/4)πd^{2}h you can substitute d/5 for h and solve for d in meters which I leave to you.

Part b) I'm assuming that the course is calculus-based because the bouyancy force is not constant with depth submerged:

The change of energy going from the equilibrium position to the submerged position (The Min W required)

ΔE = Int(from 1/3 h to h) of (A(P(z)-Patm)dz) - ρVgΔz where z is the depth from the surface.

ΔE = Aρ_{w}g * Int(from h/3 to h) of (zdz) - m_{s}gz = (πd^{2}/4)ρ_{w} g(z^{2}/2) from h/3 to h - m_{s}g(2/3 h)

***ΔE = (πd^{2}/4)ρ_{w}g(8h^{2}/18) - m_{s}g(2/3 h) (BTW, use 1000 kg/m^{3} for density of water.)

A way to do this without the integral is to realize that the value of a linearly changing force has an average equal to the average of the endpoints. This way the energy change due to work against the buoyant force will be F_{avg}Δz or 1/2 (F(h) + F(h/3)) (2h/3) = (1/2)Aρ_{w}g(h+h/3)(2/3 h) = Aρ_{w}g(8h^{2}/18) = Aρ_{w}g(4/9)h^{2}

I guess easier still is multiplying F(2/3 h)(h-h/3). These all give the same answer for the integral term.

b) I will leave this to you. The work will be due to an increase in elevation - the average bouyant force times the difference in height.

Armad K.

Thank you for your help sir! Appreciate it!09/18/20