The arranghement shown in the figure below consits of a block of mass 0.1 kg attached to a spring of force cosntant `2xx10^(-3)N//M` with the help of a pulley so that the spring remians unstretched.On releasing the block , it starts oscillating with a small amplitude . The maximum velocity of block during this motion is close to

Options

A. 72m/s
B. 71m/s
C. 70m/s
D. 69m/s

Explanation

To solve the problem step by step, we will determine the maximum velocity of the block attached to the spring when it starts oscillating after being released.

### Step 1: Identify the parameters
- Mass of the block, \( m = 0.1 \, \text{kg} \)
- Spring constant, \( k = 2 \times 10^{-3} \, \text{N/m} \)

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) for a mass-spring system is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{2 \times 10^{-3}}{0.1}} = \sqrt{0.02} = 0.1414 \, \text{rad/s} \]

### Step 3: Determine the amplitude \( A \)
When the block is released, it will stretch the spring until the spring force equals the gravitational force acting on the block. At maximum stretch (amplitude), we have:
\[ kA = mg \]
Rearranging for \( A \):
\[ A = \frac{mg}{k} \]
Substituting the known values:
\[ A = \frac{0.1 \times 9.8}{2 \times 10^{-3}} = \frac{0.98}{0.002} = 490 \, \text{m} \]

### Step 4: Calculate the maximum velocity \( V_{\text{max}} \)
The maximum velocity in simple harmonic motion is given by:
\[ V_{\text{max}} = \omega A \]
Substituting the values of \( \omega \) and \( A \):
\[ V_{\text{max}} = 0.1414 \times 490 \approx 69.3 \, \text{m/s} \]

### Conclusion
Thus, the maximum velocity of the block during its motion is approximately \( 69.3 \, \text{m/s} \).