To solve the problem of finding the extension produced by a 10N force on a spring, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension of the spring, as long as the elastic limit is not exceeded. Mathematically, this is expressed as:
\[ F = k \cdot x \]
where:
- \( F \) is the force applied,
- \( k \) is the spring constant,
- \( x \) is the extension of the spring.
First, we need to find the spring constant \( k \) using the given information. The problem states that an 8N force produces a 4cm (0.04m) extension. Using Hooke's Law, we rearrange the equation to find \( k \):
\[ k = \frac{F}{x} = \frac{8, \text{N}}{0.04, \text{m}} = 200, \text{N/m} \]
Now, we can determine the extension produced by a 10N force. We will again use Hooke's Law:
- Rearranging as \( x = \frac{F}{k} \)
- Substituting \( F = 10, \text{N} \) and \( k = 200, \text{N/m} \):
\[ x = \frac{10, \text{N}}{200, \text{N/m}} = 0.05, \text{m} \]
This equates to an extension of:
\[ x = 0.05, \text{m} = 5, \text{cm} \]
Therefore, the extension produced by a 10N force is 5 cm.
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