To solve this problem, we can use Hooke's Law, which states that the force \( F \) applied to a spring (or elastic material) is directly proportional to the extension \( x \), as long as the elastic limit is not exceeded. The mathematical representation of Hooke's Law is:
\[ F = kx \]
where:
- \( F \) is the force applied,
- \( k \) is the spring constant, and
- \( x \) is the extension.
From the information given, a force of 2 Newtons stretches the elastic material by 30mm (0.03m). We can use this to find the spring constant \( k \).
Step 1: Calculate the spring constant \( k \)
Using the values:
- \( F = 2 , \text{N} \)
- \( x = 30 , \text{mm} = 0.03 , \text{m} \)
We can plug these values into Hooke's law:
\[ 2 = k \cdot 0.03 \]
To find \( k \), we rearrange the equation:
\[ k = \frac{2}{0.03} \approx 66.67 , \text{N/m} \]
Step 2: Find the force required to stretch the material by 40mm
Now, we want to find the force required to stretch the material by 40mm (0.04 m).
Using Hooke's law again:
\[ F = kx \]
Where \( x = 40 , \text{mm} = 0.04 , \text{m} \):
\[ F = 66.67 \cdot 0.04 \]
Calculating this gives:
\[ F \approx 2.67 , \text{N} \]
Step 3: Find the additional force required
Since we previously applied a force of 2 N to stretch it to 30 mm, we need to find the additional force required to stretch it to 40 mm:
\[ \text{Additional force} = F - \text{previous force} \]
\[ \text{Additional force} = 2.67 , \text{N} - 2 , \text{N} \]
\[ \text{Additional force} = 0.67 , \text{N} \]
Conclusion
The additional force required to stretch the elastic material by 40mm is approximately 0.67 Newtons.
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