A block of ice with an area A and a height h floats in water of density `rho_(0)`. What work should be performed to submerge the ice block completely into water if density of ice is `rho_(1)` ?

Answer

To solve the problem of calculating the work required to completely submerge a block of ice in water, we can follow these steps:

### Step 1: Understand the Forces Acting on the Ice Block
When the ice block is floating, it experiences two main forces: the gravitational force acting downward and the buoyant force acting upward. The gravitational force is given by the weight of the ice, while the buoyant force is equal to the weight of the water displaced by the submerged part of the ice.

### Step 2: Calculate the Weight of the Ice Block
The weight of the ice block can be calculated using the formula:
\[ W_{ice} = \rho_1 \cdot g \cdot V_{ice} \]
where \(V_{ice} = A \cdot h\) (the volume of the ice block), so:
\[ W_{ice} = \rho_1 \cdot g \cdot (A \cdot h) \]

### Step 3: Calculate the Buoyant Force
The buoyant force acting on the ice block when it is partially submerged can be calculated using Archimedes' principle:
\[ F_{buoyant} = \rho_0 \cdot g \cdot V_{displaced} \]
where \(V_{displaced} = A \cdot s\) (the volume of water displaced), so:
\[ F_{buoyant} = \rho_0 \cdot g \cdot (A \cdot s) \]

### Step 4: Determine the Depth of Submersion
To find the depth \(s\) to which the ice block is submerged when floating, we apply the equilibrium condition (the weight of the ice equals the buoyant force):
\[ \rho_1 \cdot g \cdot (A \cdot h) = \rho_0 \cdot g \cdot (A \cdot s) \]
Cancelling \(g\) and \(A\) from both sides, we get:
\[ \rho_1 \cdot h = \rho_0 \cdot s \]
From this, we can solve for \(s\):
\[ s = \frac{\rho_1}{\rho_0} \cdot h \]

### Step 5: Calculate the Work Done to Submerge the Ice Block
When the ice block is submerged, we need to consider the work done against the buoyant force. The work done \(W\) can be calculated as:
\[ W = \int_{s}^{h} (W_{ice} - F_{buoyant}) \, dx \]
Where \(dx\) is the infinitesimal distance moved downwards. The limits of integration are from \(s\) (initially submerged depth) to \(h\) (fully submerged).

### Step 6: Substitute the Forces into the Work Equation
Substituting the expressions for \(W_{ice}\) and \(F_{buoyant}\):
\[ W = \int_{s}^{h} \left( \rho_1 \cdot g \cdot (A \cdot h) - \rho_0 \cdot g \cdot (A \cdot x) \right) \, dx \]

### Step 7: Solve the Integral
Calculating the integral:
\[ W = \rho_1 \cdot g \cdot A \cdot \left( h \cdot (h - s) \right) - \frac{1}{2} \rho_0 \cdot g \cdot A \cdot (h^2 - s^2) \]

### Step 8: Substitute for \(s\) and Simplify
Substituting \(s = \frac{\rho_1}{\rho_0} \cdot h\) into the work equation and simplifying gives us the final expression for the work done to submerge the ice block completely.

### Final Expression
The final expression for the work done can be simplified to:
\[ W = \frac{g \cdot A \cdot h^2}{2} \cdot \left( \frac{\rho_0 - \rho_1}{\rho_0} \right) \]