a force of 10n causes a spring to extend by 2mm find a, the spring constant of the spring in N/w b, the extension of the spring when 25N is applied c, the force applied that caused an extension of 5mm

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

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