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Q5) A net force \( F \) acts on a mass \( m \) and produces an acceleration \( a \). What acceleration results if a net force \( 3F \) acts on mass \( 6m \)?

A. \(\frac{a}{2}\)
B. \(8a\)
C. \(4a\)
D. \(2a\)
E. [tex]\(\frac{a}{4}\)[/tex]


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]
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