A block of mass M is suspended from a wire of length L, area of cross-section A and Young's modulus Y. The elastic potential energy stored in the wire is

Options

A. `(1)/(2) (M^(2) g^(2) L)/(AY)`
B. `(1)/(2) (Mg)/(ALY)`
C. `(1)/(2) (M^(2)g^(2)A)/(YL)`
D. `(1)/(2) (MgY)/(AL)`

Answer

To find the elastic potential energy stored in a wire when a block of mass \( M \) is suspended from it, we can follow these steps:

### Step 1: Understand the relationship between stress, strain, and Young's modulus.
Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress \( = \frac{F}{A} \) (Force per unit area)
- Strain \( = \frac{\Delta L}{L} \) (Change in length per original length)

### Step 2: Identify the forces acting on the wire.
When the block of mass \( M \) is suspended, the force \( F \) acting on the wire due to the weight of the block is:
\[ F = Mg \]
Where \( g \) is the acceleration due to gravity.

### Step 3: Express stress and strain in terms of the given quantities.
Substituting the expression for force into the stress formula:
\[ \text{Stress} = \frac{F}{A} = \frac{Mg}{A} \]
Now, substituting this into the Young's modulus equation:
\[ Y = \frac{\frac{Mg}{A}}{\frac{\Delta L}{L}} \]
Rearranging gives:
\[ \Delta L = \frac{MgL}{AY} \]

### Step 4: Find the spring constant \( k \) of the wire.
The spring constant \( k \) can be related to Young's modulus and the dimensions of the wire:
\[ k = \frac{AY}{L} \]

### Step 5: Calculate the elastic potential energy stored in the wire.
The elastic potential energy \( U \) stored in a spring (or wire, in this case) is given by:
\[ U = \frac{1}{2} k x^2 \]
Substituting for \( k \) and \( x \) (where \( x = \Delta L \)):
\[ U = \frac{1}{2} \left( \frac{AY}{L} \right) \left( \frac{MgL}{AY} \right)^2 \]
Simplifying this expression:
\[ U = \frac{1}{2} \cdot \frac{AY}{L} \cdot \frac{M^2g^2L^2}{A^2Y^2} \]
\[ U = \frac{1}{2} \cdot \frac{M^2g^2L}{AY} \]

### Final Result:
The elastic potential energy stored in the wire is:
\[ U = \frac{M^2g^2L}{2AY} \]

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