A block of mass m is attached with a spring in vertical plane as shown in the figure. If initially spring is in its natural length and the block is released from rest, then maximum extension in the spring will be

Options

A. `2mg/k`
B. `4mg/k`
C. `mg/k`
D. `mg/2k`

Answer

To find the maximum extension of the spring when a block of mass \( m \) is attached to it and released from rest, we can follow these steps:

### Step 1: Understand the system
The block is initially at rest, and the spring is at its natural length. When the block is released, it will fall under the influence of gravity, and the spring will stretch as the block moves downward.

### Step 2: Define the energies involved
At the maximum extension \( x \) of the spring, the gravitational potential energy lost by the block will be equal to the elastic potential energy stored in the spring.

- The gravitational potential energy lost by the block when it falls a distance \( x \) is given by:
\[ E_i = mgh = mgx \]

- The elastic potential energy stored in the spring when it is stretched by \( x \) is given by:
\[ E_f = \frac{1}{2} k x^2 \]

### Step 3: Set up the energy conservation equation
Since energy is conserved, we can equate the initial gravitational potential energy to the final elastic potential energy:
\[ mgx = \frac{1}{2} k x^2 \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ \frac{1}{2} k x^2 - mgx = 0 \]
This is a quadratic equation in \( x \).

### Step 5: Factor the equation
Factoring out \( x \) gives:
\[ x \left( \frac{1}{2} k x - mg \right) = 0 \]
This gives us two solutions: \( x = 0 \) (the trivial solution) and:
\[ \frac{1}{2} k x - mg = 0 \Rightarrow kx = 2mg \Rightarrow x = \frac{2mg}{k} \]

### Step 6: Conclusion
Thus, the maximum extension \( x \) of the spring is:
\[ x = \frac{2mg}{k} \]

### Final Answer
The maximum extension in the spring will be \( \frac{2mg}{k} \).

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