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. Mg/k
B. 2Mg/k
C. 4Mg/k
D. Mg/2k

Answer

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

### Step 1: Understand the System
When the block is released, it will fall under the influence of gravity, and the spring will stretch as a result. Initially, the spring is unstretched, and the block has gravitational potential energy.

### Step 2: Define the Variables
- Let \( k \) be the spring constant.
- Let \( M \) be the mass of the block.
- Let \( h \) be the maximum extension of the spring.

### Step 3: Energy Conservation Principle
At the maximum extension \( h \), the block momentarily comes to rest, meaning all its gravitational potential energy has been converted into spring potential energy.

The gravitational potential energy (P.E.) lost by the block as it falls a distance \( h \) is given by:
\[ \text{P.E.}_{\text{gravity}} = Mgh \]

The potential energy stored in the spring when it is stretched by \( h \) is given by:
\[ \text{P.E.}_{\text{spring}} = \frac{1}{2} k h^2 \]

### Step 4: Set Up the Energy Conservation Equation
According to the conservation of energy:
\[ \text{P.E.}_{\text{gravity}} = \text{P.E.}_{\text{spring}} \]
Substituting the expressions for potential energy:
\[ Mgh = \frac{1}{2} k h^2 \]

### Step 5: Rearrange the Equation
Rearranging the equation gives:
\[ \frac{1}{2} k h^2 - Mgh = 0 \]
This is a quadratic equation in terms of \( h \):
\[ \frac{1}{2} k h^2 - Mgh = 0 \]

### Step 6: Solve the Quadratic Equation
Factoring out \( h \):
\[ h \left( \frac{1}{2} k h - Mg \right) = 0 \]
This gives us two solutions: \( h = 0 \) (the trivial solution) and:
\[ \frac{1}{2} k h - Mg = 0 \implies k h = 2Mg \implies h = \frac{2Mg}{k} \]

### Step 7: Conclusion
Thus, the maximum extension \( h \) in the spring is:
\[ h = \frac{2Mg}{k} \]