Question

Flotation device is in the shape of a right cylinder, with a height of $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{m}$and a face area of $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$on top and bottom, and its density is localid="1657554653816" $\mathbf{0}\mathbf{.}\mathbf{400}$ times that of fresh water. It is initially held fully submerged in fresh water, with its top face at the water surface. Then it is allowed to ascend gradually until it begins to float. How much work does the buoyant force do on the device during the ascent?

Short answer

The work done by buoyant force on the device during ascent is$4.11\mathrm{kJ}$

Step-by-step solution

The given data

i) The height of the cylinder is$\mathrm{h}=0.500\mathrm{m}$

ii) The face area of the cylinder is$\mathrm{A}=4.00{\mathrm{m}}^2$

iii) The density of the cylinder is${\mathrm{\rho }}_{\mathrm{c}}-0.400{\mathrm{\rho }}_{\mathrm{z}}$

Understanding the concept of Archimedes Principle

We can use Archimedes’ principle to write an expression for force of buoyancy. Then taking its integration for y for its displacement during ascent, we can find the work done by the force of buoyancy. We can use the relation of pressure at depth to writing its displacement in terms of h. By putting it in the expression for W, we can get the magnitude of work done by the buoyant force on the device during ascent.

Formulae:

Downward force due to gravity, ${\mathrm{F}}_{\mathrm{b}}={\mathrm{m}}_{\mathrm{f}}\mathrm{g}$ (i)

Work done by a body, $\mathrm{W}=\int {\mathrm{F}}_{\mathrm{b}}\mathrm{dy}$(ii)

Density of a substance, $\mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}$(iii)

Pressure applied on a body, $\mathrm{p}=\mathrm{pgh}$(iv)

Calculation of work done by buoyant force

When a cylinder submerged in a fluid, a buoyant force from the surrounding fluid acts on the cylinder in the upward direction, and it is equal to the weight of the fluid that has been displaced by the body.

According to Archimedes’ principle using equations (i) & (iii), we get

${\mathrm{F}}_{\mathrm{b}}={\mathrm{\rho }}_{\mathrm{w}}{\mathrm{V}}_{\mathrm{c}}\mathrm{g}$

$={\mathrm{\rho }}_{\mathrm{w}}\mathrm{Ag}(-\mathrm{y})$

We can consider the origin at the surface of the water. Initially, the full cylinder is submerged, and then it is allowed to ascent. Hence, the integral has lower limit as $-\mathrm{h}$and upper limit as ${\mathrm{y}}_{\mathrm{f}}$. The work done by the buoyant force from equation (ii) is given by:

$\mathrm{W}={\int }_{-\mathrm{h}}^{{\mathrm{y}}_{\mathrm{t}}} {\mathrm{F}}_{\mathrm{b}}\mathrm{dy}$

$={\int }_{-\mathrm{h}}^{{\mathrm{y}}_{\mathrm{t}}} {\mathrm{\rho }}_{\mathrm{w}}\mathrm{Ag}(-\mathrm{y})\mathrm{dy}$

$={\mathrm{\rho }}_{\mathrm{w}}\mathrm{Ag}{\left(\frac{-{\mathrm{y}}^2}{2}\right)}_{\mathrm{n}}^{{\mathrm{y}}_{\mathrm{f}}}$

$=\frac{1}{2}{\mathrm{\rho }}_{\mathrm{w}}\mathrm{Ag}\left(-{\mathrm{y}}_{\mathrm{f}}^2+{\mathrm{h}}^2\right)$

The height of the cylinder is $-\mathrm{h}$, and the height of the water is ${\mathrm{y}}_{\mathrm{f}}$. The pressure at depth relation using equation (iv) is given by:

${\text{h}}_{\mathrm{c}}{\mathrm{\rho }}_{\mathrm{c}}\mathrm{g}={\mathrm{h}}_{\mathrm{w}}{\mathrm{P}}_{\mathrm{w}}\mathrm{g}$

$-{\mathrm{h\rho }}_{\mathrm{c}}={\mathrm{y}}_{\mathrm{f}}{\mathrm{\rho }}_{\mathrm{u}}$

$\frac{{\mathrm{y}}_{\mathrm{p}}}{\mathrm{h}}=\frac{-{\mathrm{\rho }}_{\mathrm{c}}}{{\mathrm{\rho }}_{\mathrm{w}}}$

${\mathrm{y}}_{\mathrm{f}}=\frac{-0.400{\mathrm{\rho }}_{\mathrm{w}}}{{\mathrm{\rho }}_{\mathrm{w}}}\mathrm{h}$

$=-0.400\mathrm{h}$

Then after putting the given values equation (2) becomes:

$W=\frac{1}{2}{\rho }_{w}Ag\left[-(-0.400h{)}^2+h^2\right]$

$=\frac{1}{2}{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{Agh}}^2\left[1-(-0.400{)}^2\right]$

$=\frac{1}{2}\times 1000\frac{kg}{m^3}\times 4.00m^2\times 9.8m/s^2\times (0.500m{)}^2\left[1-(-0.400{)}^2\right]$

$=4.12\mathrm{kJ}$

Hence, the value of work done is$4.12\mathrm{kJ}$