Question

A certain spring has a resting length of 0.08 m (8 cm) when it is not stretched. The spring is stretched to a length of 0.10 m (10 cm), 0.02 m beyond its natural length. Holding it in this stretched position requires a force of 20 N. Question: What is the force function for this spring? Answer: Hooke's Law says that the force required to maintain the spring in a stretched position is the linear function F(x) = kx. Find the spring constant, k, for the spring described above, and enter that number as your final answer.

Answer 1

Answer:

The spring constant is   $k =1000 \ N/m$

Step-by-step explanation:

From the question we are told that

     The resting length is $x = 0.08 \ m$

        The extension of spring is  $e = 0.02 \ m$

      The new length of spring is  $l = 0.10 \ m$

       The force is  $F = 20 \ N$

The spring constant is

         $k = \frac{F}{e}$

substituting values

         $k = \frac{20}{0.02}$

        $k =1000 \ N/m$