A vertical spring with force constant `k` is fixed on a table. A ball of mass `m` at a height `h` above the free upper end of the spring falls vertically on the spring , so that the spring is compressed by a distance `d`. The net work done in the process is

Options

A. `mg(h-d)+(1)/(2)Kd^(2)`
B. `mg(h+d)+(1)/(2)Kd^(2)`
C. `mg(h+d)-(1)/(2)Kd^(2)`
D. `mg(h-d)-(1)/(2)Kd^(2)`

Explanation

To solve the problem, we need to calculate the net work done when a ball of mass \( m \) falls from a height \( h \) and compresses a vertical spring by a distance \( d \).

### Step-by-step Solution:

1. **Identify Forces Acting on the Ball:**
- The gravitational force acting on the ball is \( F_g = mg \), where \( g \) is the acceleration due to gravity.
- As the ball compresses the spring, the spring exerts an upward force given by Hooke's Law: \( F_s = kx \), where \( x \) is the compression of the spring.

2. **Calculate Work Done by Gravity:**
- The ball falls a total distance of \( h + d \) (it falls \( h \) to reach the spring and then compresses the spring by \( d \)).
- The work done by gravity \( W_g \) is positive because the force and displacement are in the same direction:
\[ W_g = F_g \cdot \text{displacement} = mg(h + d) \]

3. **Calculate Work Done by the Spring:**
- The work done by the spring \( W_s \) is negative because the spring force acts in the opposite direction to the displacement of the ball:
\[ W_s = -\frac{1}{2} k d^2 \]
- This is derived from the work done on a spring, which is \( \frac{1}{2} k x^2 \) for compression \( x \).

4. **Calculate Net Work Done:**
- The net work done \( W_{net} \) is the sum of the work done by gravity and the work done by the spring:
\[ W_{net} = W_g + W_s \]
- Substituting the expressions we found:
\[ W_{net} = mg(h + d) - \frac{1}{2} k d^2 \]

### Final Answer:
The net work done in the process is:
\[ W_{net} = mg(h + d) - \frac{1}{2} k d^2 \]