A smooth uniform, string of natural length `l`, cross-sectional area `A` and Young’s modulus `Y` is pulled along its length by a force `F` on a horizontal surface. Find the elastic potential energy stored in the string.

Explanation

To find the elastic potential energy stored in a smooth uniform string of natural length \( l \), cross-sectional area \( A \), and Young’s modulus \( Y \) when it is pulled by a force \( F \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a string with natural length \( l \), cross-sectional area \( A \), and Young's modulus \( Y \). When a force \( F \) is applied, the string stretches, and we need to find the elastic potential energy stored in it.

2. **Tension in the String**:
Consider a point at a distance \( x \) from one end of the string. The tension \( T \) at this point due to the force \( F \) can be expressed as:
\[ T = \frac{F \cdot x}{l} \]

3. **Stress in the String**:
The stress \( \sigma \) at this point can be calculated using the formula:
\[ \sigma = \frac{T}{A} = \frac{F \cdot x}{A \cdot l} \]

4. **Elastic Potential Energy Formula**:
The elastic potential energy \( U \) stored in the string can be calculated using the formula:
\[ U = \frac{1}{2} Y \int \sigma^2 \, dv \]
where \( dv \) is the differential volume element. For a length \( dx \) of the string, \( dv = A \, dx \).

5. **Substituting Stress into the Energy Formula**:
Substitute the expression for stress \( \sigma \) into the energy formula:
\[ U = \frac{1}{2} Y \int_0^l \left(\frac{F \cdot x}{A \cdot l}\right)^2 A \, dx \]

6. **Simplifying the Integral**:
This simplifies to:
\[ U = \frac{1}{2} Y \cdot \frac{F^2}{A^2 l^2} \int_0^l x^2 \, dx \]
The integral \( \int_0^l x^2 \, dx \) can be calculated as:
\[ \int_0^l x^2 \, dx = \frac{l^3}{3} \]

7. **Final Calculation**:
Plugging this back into the equation for \( U \):
\[ U = \frac{1}{2} Y \cdot \frac{F^2}{A^2 l^2} \cdot \frac{l^3}{3} \]
Simplifying this gives:
\[ U = \frac{1}{6} \cdot \frac{F^2}{A Y} \]

### Final Result:
The elastic potential energy stored in the string is:
\[ U = \frac{1}{6} \cdot \frac{F^2}{A Y} \]