Question

An 8N force stretches a spring to the 20cm mark rule. If the extension on the spring is 4cm, calculate the; extension produced by a 10N force

Answer 1

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:

1. Rearranging as \( x = \frac{F}{k} \)
2. 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**.