To solve this problem, we can use Hooke's law, which states that \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the stretch of the spring.
First, we can find the spring constant \( k \) using the information given for the 100 newtons force:
\[ F = kx \implies k = \frac{F}{x} \]
For the force of 100 newtons stretching the spring 0.25 meters, we have:
\[ k = \frac{100 , \text{N}}{0.25 , \text{m}} = 400 , \text{N/m} \]
Next, we will use this spring constant to find out how far the spring stretches with a force of 80 newtons:
\[ F = kx \implies x = \frac{F}{k} \]
Substituting the values for \( F = 80 \) N and \( k = 400 \) N/m:
\[ x = \frac{80 , \text{N}}{400 , \text{N/m}} = 0.2 , \text{m} \]
Thus, a force of 80 newtons will stretch the spring 0.2 meters.
The correct response is 0.2 meters.