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Sagot :
Let's solve this step-by-step using fundamental principles from classical mechanics:
### Step-by-Step Solution:
1. Understanding the Given Information:
- A net force \( F \) acts on mass \( m \) and produces an acceleration \( a \).
2. Using Newton's Second Law:
- Newton's second law states: \( F = m \cdot a \).
3. Deriving the Acceleration from the Given Information:
- From the relationship \( F = m \cdot a \), we can express the acceleration \( a \) as:
[tex]\[ a = \frac{F}{m} \][/tex]
4. New Conditions:
- A new net force \( 3F \) acts on a new mass \( 6m \).
5. Calculating the New Acceleration:
- Using Newton's second law again for the new force and mass:
[tex]\[ \text{New acceleration}, a' = \frac{\text{New force}}{\text{New mass}} = \frac{3F}{6m} \][/tex]
6. Simplifying the Expression:
- Simplify the fraction:
[tex]\[ a' = \frac{3F}{6m} = \frac{1}{2} \cdot \frac{F}{m} \][/tex]
7. Relating to the Original Acceleration:
- Recall from step 3 that \( a = \frac{F}{m} \). Substituting this in, we get:
[tex]\[ a' = \frac{1}{2} \cdot a \][/tex]
So, the resulting acceleration \( a' \) when a net force of \( 3F \) acts on a mass of \( 6m \) is half the original acceleration.
### Conclusion:
The acceleration that results if a net force of \( 3F \) acts on a mass of \( 6m \) is:
[tex]\[ a' = \frac{a}{2} \][/tex]
Hence, the correct answer is:
a. [tex]\( \frac{a}{2} \)[/tex]
### Step-by-Step Solution:
1. Understanding the Given Information:
- A net force \( F \) acts on mass \( m \) and produces an acceleration \( a \).
2. Using Newton's Second Law:
- Newton's second law states: \( F = m \cdot a \).
3. Deriving the Acceleration from the Given Information:
- From the relationship \( F = m \cdot a \), we can express the acceleration \( a \) as:
[tex]\[ a = \frac{F}{m} \][/tex]
4. New Conditions:
- A new net force \( 3F \) acts on a new mass \( 6m \).
5. Calculating the New Acceleration:
- Using Newton's second law again for the new force and mass:
[tex]\[ \text{New acceleration}, a' = \frac{\text{New force}}{\text{New mass}} = \frac{3F}{6m} \][/tex]
6. Simplifying the Expression:
- Simplify the fraction:
[tex]\[ a' = \frac{3F}{6m} = \frac{1}{2} \cdot \frac{F}{m} \][/tex]
7. Relating to the Original Acceleration:
- Recall from step 3 that \( a = \frac{F}{m} \). Substituting this in, we get:
[tex]\[ a' = \frac{1}{2} \cdot a \][/tex]
So, the resulting acceleration \( a' \) when a net force of \( 3F \) acts on a mass of \( 6m \) is half the original acceleration.
### Conclusion:
The acceleration that results if a net force of \( 3F \) acts on a mass of \( 6m \) is:
[tex]\[ a' = \frac{a}{2} \][/tex]
Hence, the correct answer is:
a. [tex]\( \frac{a}{2} \)[/tex]
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