# A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm?

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gsenviro | Certified Educator

Force = 30 N, elongation = 15 cm-12 cm = 3 cm

Spring force, F = kx, where k is spring constant and x is elongation,

thus k = F/x = 30 N/ 3 cm = 10 N/cm

Work done = 1/2 kx^2

In this case, x = 20 cm - 12 cm = 8 cm

thus work done, W = 1/2 . 10 N/cm . 8 cm . 8 cm = **320 N.cm = 3.2 N m = 3.2 J**