A spring of natural length `l` is compressed vertically downward against the floor so that its compressed lengtg becomes `(1)/(2)`. On releasing, the spring attins its natural length. If `k` is the siffness constant of spring, then the work done by the spring on the floor is

Options

A. `zero`
B. `1/2kl^(2)`
C. `(1)/(2)k((l^(2))/(2))`
D. `kl^(2)`

Explanation

To solve the problem, we need to determine the work done by the spring on the floor when it is compressed and then released.

### Step-by-Step Solution:

1. **Understanding the Spring Compression**:
- The spring has a natural length `l`. When compressed, its length becomes `l/2`. This means the spring is compressed by a distance of:
\[ \Delta x = l - \frac{l}{2} = \frac{l}{2} \]

2. **Calculating the Force Exerted by the Spring**:
- The force exerted by the spring when it is compressed can be calculated using Hooke's Law:
\[ F = k \Delta x \]
- Substituting the value of \(\Delta x\):
\[ F = k \cdot \frac{l}{2} \]

3. **Understanding Work Done**:
- The work done by the spring when it returns to its natural length can be calculated using the formula for work done by a variable force:
\[ W = \frac{1}{2} k (\Delta x)^2 \]
- Substituting \(\Delta x = \frac{l}{2}\):
\[ W = \frac{1}{2} k \left(\frac{l}{2}\right)^2 = \frac{1}{2} k \cdot \frac{l^2}{4} = \frac{kl^2}{8} \]

4. **Considering the Work Done on the Floor**:
- However, the question specifically asks for the work done by the spring on the floor. Since the floor does not move (it is fixed), the displacement \(s\) of the floor is 0.
- Therefore, the work done by the spring on the floor is:
\[ W_{\text{floor}} = F \cdot s = F \cdot 0 = 0 \]

5. **Conclusion**:
- The work done by the spring on the floor is zero because there is no displacement of the floor.

### Final Answer:
The work done by the spring on the floor is **0**.