A spring of unstretched length l has a mass m with one end fixed to a rigid support.Assuming spring to be made of a uniform wire,the kinetic energy possessed by it is its free end is pulled with uniform velocity v is :

Options

A. `(1)/(2)mv^(2)`
B. `mv^(2)`
C. `(1)/(3)mv^(2)`
D. `(1)/(6)mv^(2)`

Answer

To solve the problem, we need to find the kinetic energy possessed by a spring when its free end is pulled with a uniform velocity \( v \).

### Step-by-step Solution:

1. **Understanding the Spring's Properties**:
- The spring has an unstretched length \( l \) and a mass \( m \).
- The spring is assumed to be made of a uniform wire, which means its mass is distributed evenly along its length.

2. **Calculating Linear Mass Density**:
- The linear mass density \( \lambda \) of the spring can be calculated as:
\[ \lambda = \frac{m}{l} \]
- This gives us the mass per unit length of the spring.

3. **Considering an Element of the Spring**:
- Let’s consider a small element \( dx \) of the spring at a distance \( x \) from the fixed end.
- The mass of this small element \( dm \) can be expressed as:
\[ dm = \lambda \, dx = \frac{m}{l} \, dx \]

4. **Finding the Velocity of the Element**:
- When the free end of the spring is pulled with a velocity \( v \), the velocity of the element at distance \( x \) from the fixed end can be determined by the proportionality of the distance:
\[ v_x = \frac{x}{l} v \]
- This means that the velocity of each element varies linearly from \( 0 \) (at the fixed end) to \( v \) (at the free end).

5. **Calculating Kinetic Energy of the Element**:
- The kinetic energy \( dK \) of the small element \( dm \) is given by:
\[ dK = \frac{1}{2} dm \, v_x^2 \]
- Substituting \( dm \) and \( v_x \):
\[ dK = \frac{1}{2} \left(\frac{m}{l} \, dx\right) \left(\frac{x}{l} v\right)^2 \]
\[ dK = \frac{1}{2} \frac{m}{l} \frac{x^2 v^2}{l^2} \, dx \]
\[ dK = \frac{mv^2}{2l^3} x^2 \, dx \]

6. **Integrating to Find Total Kinetic Energy**:
- To find the total kinetic energy \( K \) of the entire spring, we integrate \( dK \) from \( 0 \) to \( l \):
\[ K = \int_0^l \frac{mv^2}{2l^3} x^2 \, dx \]
- Evaluating the integral:
\[ K = \frac{mv^2}{2l^3} \left[ \frac{x^3}{3} \right]_0^l = \frac{mv^2}{2l^3} \cdot \frac{l^3}{3} = \frac{mv^2}{6} \]

### Final Result:
The total kinetic energy possessed by the spring when its free end is pulled with a uniform velocity \( v \) is:
\[ K = \frac{mv^2}{6} \]