A block of density `2000kg//m^3` and mass 10kg is suspended by a spring stiffness `100N//m`. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density `1000kg//m^3`. If the block is in equilibrium position.

Options

A. the elongation of the spring is 1 cm
B. the magnitude of buoyant force acting on the block is 50 N
C. the spring potential energy is 12.5 J
D. magnitude of spring force on the block is greater than the weight of the block

Answer

To solve the problem step by step, we will analyze the forces acting on the block and apply the principles of equilibrium.

### Step 1: Identify the forces acting on the block
When the block is submerged in the liquid and in equilibrium, three forces act on it:
1. The weight of the block (downward).
2. The buoyant force acting on the block (upward).
3. The force exerted by the spring (upward).

### Step 2: Calculate the weight of the block
The weight of the block can be calculated using the formula:
\[ W = m \cdot g \]
Where:
- \( m = 10 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Calculate the buoyant force
The buoyant force can be calculated using Archimedes' principle:
\[ F_b = V_b \cdot \rho_l \cdot g \]
Where:
- \( V_b \) is the volume of the block.
- \( \rho_l = 1000 \, \text{kg/m}^3 \) (density of the liquid).

To find the volume of the block, we can use its density:
\[ V_b = \frac{m}{\rho_b} = \frac{10 \, \text{kg}}{2000 \, \text{kg/m}^3} = 0.005 \, \text{m}^3 \]

Now, substituting in the buoyant force formula:
\[ F_b = 0.005 \, \text{m}^3 \cdot 1000 \, \text{kg/m}^3 \cdot 10 \, \text{m/s}^2 = 50 \, \text{N} \]

### Step 4: Calculate the force exerted by the spring
The force exerted by the spring can be expressed as:
\[ F_s = k \cdot x \]
Where:
- \( k = 100 \, \text{N/m} \) (spring constant)
- \( x \) is the extension of the spring.

### Step 5: Apply equilibrium condition
In equilibrium, the sum of upward forces equals the sum of downward forces:
\[ F_b + F_s = W \]
Substituting the known values:
\[ 50 \, \text{N} + (100 \, \text{N/m} \cdot x) = 100 \, \text{N} \]

### Step 6: Solve for \( x \)
Rearranging the equation:
\[ 100 \, \text{N/m} \cdot x = 100 \, \text{N} - 50 \, \text{N} \]
\[ 100 \, \text{N/m} \cdot x = 50 \, \text{N} \]
\[ x = \frac{50 \, \text{N}}{100 \, \text{N/m}} = 0.5 \, \text{m} \]

### Step 7: Calculate the potential energy stored in the spring
The potential energy stored in the spring can be calculated using:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values:
\[ PE = \frac{1}{2} \cdot 100 \, \text{N/m} \cdot (0.5 \, \text{m})^2 = \frac{1}{2} \cdot 100 \cdot 0.25 = 12.5 \, \text{J} \]

### Final Results
1. The buoyant force acting on the block is \( 50 \, \text{N} \).
2. The potential energy stored in the spring is \( 12.5 \, \text{J} \).