A body of mass 100 g is attached to a hanging spring force constant is `10 N//m`. The body is lifted until the spring is in its unstretched state and then released. Calculate the speed of the body when it strikes the table `15 cm` below the release point .

Options

A. `1 m//s`
B. `0.866 m//s`
C. `0.225 m//s`
D. `1.5 m//s`

Explanation

To solve the problem, we will use the principle of conservation of energy. The energy in the system will be conserved as the body falls and strikes the table.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the body, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Spring constant, \( k = 10 \, \text{N/m} \)
- Distance fallen, \( h = 15 \, \text{cm} = 0.15 \, \text{m} \)

2. **Initial Energy:**
- When the body is lifted to the unstretched position of the spring, the initial energy of the system is zero because the spring is not stretched and the body is at rest.

3. **Final Energy:**
- When the body strikes the table, the final energy consists of:
- Kinetic energy of the body: \( KE = \frac{1}{2} mv^2 \)
- Potential energy of the body at the height of the table: \( PE = -mgh \) (negative because it is below the reference point)
- Elastic potential energy stored in the spring when it is stretched: \( PE_{spring} = \frac{1}{2} kx^2 \) where \( x = 0.15 \, \text{m} \)

4. **Setting Up the Energy Conservation Equation:**
\[ \text{Initial Energy} = \text{Final Energy} \]
\[ 0 = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 - mgh \]

5. **Substituting Values:**
- Substitute \( k = 10 \, \text{N/m} \), \( x = 0.15 \, \text{m} \), \( m = 0.1 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \), and \( h = 0.15 \, \text{m} \):
\[ 0 = \frac{1}{2} \cdot 10 \cdot (0.15)^2 + \frac{1}{2} \cdot 0.1 \cdot v^2 - 0.1 \cdot 10 \cdot 0.15 \]

6. **Calculating Each Term:**
- Calculate \( \frac{1}{2} \cdot 10 \cdot (0.15)^2 \):
\[ = 0.5 \cdot 10 \cdot 0.0225 = 0.1125 \, \text{J} \]
- Calculate \( -0.1 \cdot 10 \cdot 0.15 \):
\[ = -0.1 \cdot 10 \cdot 0.15 = -0.15 \, \text{J} \]

7. **Rearranging the Equation:**
\[ 0 = 0.1125 + \frac{1}{2} \cdot 0.1 \cdot v^2 - 0.15 \]
\[ \frac{1}{2} \cdot 0.1 \cdot v^2 = 0.15 - 0.1125 \]
\[ \frac{1}{2} \cdot 0.1 \cdot v^2 = 0.0375 \]

8. **Solving for \( v^2 \):**
\[ 0.1 \cdot v^2 = 0.075 \]
\[ v^2 = \frac{0.075}{0.1} = 0.75 \]

9. **Finding \( v \):**
\[ v = \sqrt{0.75} \approx 0.866 \, \text{m/s} \]

### Final Answer:
The speed of the body when it strikes the table is approximately **0.866 m/s**.