Question

A force of 10 pounds is required to stretch a spring 4 inches beyond its natural length. Assuming Hooke's law applies, how much work is done in stretching the spring from its natural length to 6 inches beyond its natural length?

Answer 1

The work done is 5.084 J

Explanation:

From Hooke's law of elasticity,

F = ke

F/e = k

F1/e1 = F2/e2

F2 = F1e2/e1

F1 = 10 lbf, e2 = 6 in, e1 = 4 in

F2 = 10×6/4 = 15 lbf

Work done (W) = 1/2F2e2

F2 = 15 lbf = 15×4.4482 = 66.723 N

e2 = 6 in = 6×0.0254 = 0.1524 m

W = 1/2×66.723×0.1524 = 5.084 J