A block of mass m slides m slides down a smooth incline of angle of inclination `theta`. The force acting on it are N (normal reaction). And mg (weight). The accelration of the block is `alpha`. Two students analyze the situation and apply Newton a second law. `{:("First student","Second student"),(mg sin theta=ma,mg-Ncos theta=ma sin theta):}`

Options

A. Only first student is correct
B. Only second student is correct
C. Both students are correct
D. Both students are wrong.

Answer

To solve the problem, we will analyze the forces acting on the block sliding down the incline and apply Newton's second law.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block, \( mg \), acts vertically downward.
- The normal force, \( N \), acts perpendicular to the surface of the incline.

2. **Resolve the Weight into Components:**
- The weight can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

3. **Apply Newton's Second Law in the Direction of the Incline:**
- The net force acting on the block along the incline is \( mg \sin \theta \).
- According to Newton's second law, this net force equals the mass times the acceleration of the block:
\[ mg \sin \theta = m \alpha \]
- This equation corresponds to the first student's analysis.

4. **Apply Newton's Second Law Perpendicular to the Incline:**
- The forces acting perpendicular to the incline are the normal force \( N \) and the perpendicular component of the weight \( mg \cos \theta \).
- The net force in this direction must equal the mass times the vertical component of the acceleration:
\[ mg - N \cos \theta = m \alpha \sin \theta \]
- This equation corresponds to the second student's analysis.

5. **Conclusion:**
- From the above analysis, both equations derived by the students are correct:
- First student's equation: \( mg \sin \theta = m \alpha \)
- Second student's equation: \( mg - N \cos \theta = m \alpha \sin \theta \)

### Final Result:
Both students are correct in their analyses.

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