When you stretch a spring 20 cm past its natural length, it exerts a force of 8

N. What is the spring constant of this spring?

40 N/m

Explanation:

F = - kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = - k (0.2m)

-k = 8N / 0.2 m

k = - 40 N/m