a. Use Hooke's law to find the spring constant k :
F = kx ⇒ 10 N = k (2 mm) = k (0.002 m) ⇒ k = 5000 N/m
b. Use Hooke's law again to find the extension x given a force of 25 N :
F = kx ⇒ 25 N = (5000 N/m) x ⇒ x = 0.005 m = 5 mm
c. We end up with the same force as in part (b) :
F = kx ⇒ F = (5000 N/m) (5 mm) = (5000 N/m) (0.005 m) = 25 N