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This problem involves the analysis of forces acting on a block attached to a spring and is best tackled by using Newton's Second Law and Hooke's Law.

### Given:
- **Spring constant**: \( k_0 \)
- **Mass of the block**: \( m_0 \)
- **Displacement of the spring from equilibrium**: \( h_0 \)

### Problem Description:
A block of mass \( m_0 \) is attached to a light spring (spring constant \( k_0 \)) and pulled downward by a distance \( h_0 \), and then released. We need to find the **acceleration of the block immediately after it is released**.

### Forces Acting on the Block:
1. **Spring Force (Upward)**: 
   - Hooke's law tells us that the spring force \( F_s \) is proportional to the displacement from equilibrium.
   - When the spring is pulled down by \( h_0 \), the restoring force is:
     \[
     F_s = k_0 h_0
     \]
   - This force acts **upward**.

2. **Gravitational Force (Downward)**:
   - The gravitational force acting on the block is:
     \[
     F_g = m_0 g
     \]
   - This force acts **downward**.

### Net Force and Acceleration:
The block is released from rest, so we need to calculate the net force acting on it just after it is released.

- **Net Force** (\( F_{\text{net}} \)):
  \[
  F_{\text{net}} = F_s - F_g = k_0 h_0 - m_0 g
  \]
  
- **Using Newton’s Second Law** (\( F = m a \)):
  The net force \( F_{\text{net}} \) produces an acceleration \( a \):
  \[
  F_{\text{net}} = m_0 a
  \]
  Therefore:
  \[
  m_0 a = k_0 h_0 - m_0 g
  \]
  
- **Solving for \( a \)**:
  \[
  a = \frac{k_0 h_0 - m_0 g}{m_0}
  \]

### Answer:
The correct answer is:

\[
\boxed{D: \frac{k_0 h_0 - m_0 g}{m_0}}
\]

### Explanation for Each Choice:
- **(A) \( g \)**: This would be correct if the only force acting was gravity, but here we also have the spring force.
- **(B) \( \frac{k_0 h_0}{m_0} \)**: This only considers the spring force without accounting for gravity.
- **(C) \( \frac{k_0 h_0 + m_0 g}{m_0} \)**: This incorrectly adds the gravitational force, suggesting both forces act in the same direction, which they do not.
- **(D) \( \frac{k_0 h_0 - m_0 g}{m_0} \)**: This correctly accounts for both the spring force and the gravitational force acting in opposite directions.

Would you like further clarification on any part of the solution?

### Follow-Up Questions:
1. How does Hooke's Law describe the force exerted by a spring?
2. What are the conditions under which the block would be in equilibrium?
3. How would the acceleration change if the spring constant \( k_0 \) were increased?
4. What is the role of gravitational force in determining the net acceleration in this scenario?
5. How would the net force change if the displacement \( h_0 \) were doubled?

### Tip:
Always consider both the restoring force (spring force) and the gravitational force when dealing with spring systems under gravity. The direction of forces is crucial in determining net force and acceleration.

A light spring with spring constant k0 hangs vertically. A block of mass m0 is attached to the spring and pulled downward until the spring is displaced by a distance h0. The block is then released from rest and accelerates upward. What is the acceleration of the block immediately after it is released?

Learn how to calculate the acceleration of a block attached to a spring after it is released. This problem involves Newton's Second Law and Hooke's Law, utilizing key formulas to determine the net force and acceleration. Suitable for high school physics students in Grades 11-12.

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