An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is

Options

A. (a) `(4Mg)/(k)`
B. (b) `(2Mg)/(k)`
C. (c) `(Mg)/(k)`
D. (d) `(Mg)/(2k)`

Explanation

To find the maximum extension in the spring when a block of mass \( M \) is attached to it and released from rest, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Block
When the block is released, two main forces act on it:
- The gravitational force \( F_g = Mg \) acting downwards.
- The spring force \( F_s = kx \) acting upwards when the spring is stretched by an amount \( x \).

### Step 2: Set Up the Energy Conservation Equation
At the maximum extension of the spring, the kinetic energy of the block will be zero (it momentarily stops). Thus, the potential energy lost due to gravity will equal the potential energy stored in the spring.

The work done by gravity as the block moves down by a distance \( x \) is:
\[ W_g = -Mgx \]
The potential energy stored in the spring when it is stretched by \( x \) is:
\[ W_s = \frac{1}{2} k x^2 \]

### Step 3: Apply Conservation of Energy
According to the conservation of energy:
\[ \text{Work done by gravity} + \text{Work done by spring} = 0 \]
This gives us:
\[ -Mgx + \frac{1}{2} k x^2 = 0 \]

### Step 4: Rearrange the Equation
Rearranging the equation leads to:
\[ \frac{1}{2} k x^2 = Mgx \]

### Step 5: Solve for \( x \)
To solve for \( x \), we can rearrange the equation:
\[ \frac{1}{2} k x^2 - Mgx = 0 \]
Factoring out \( x \):
\[ x \left( \frac{1}{2} k x - Mg \right) = 0 \]
This gives us two solutions:
1. \( x = 0 \) (the trivial solution)
2. \( \frac{1}{2} k x - Mg = 0 \) which simplifies to:
\[ \frac{1}{2} k x = Mg \quad \Rightarrow \quad x = \frac{2Mg}{k} \]

### Conclusion
Thus, the maximum extension in the spring is:
\[ x = \frac{2Mg}{k} \]