A force of 18 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 14 in. beyond its natural length?

Explanation:

Given

Force=18lb

extension=8in

Using Hooke's law to get the spring constant(k)

F=ke

Then,

K=f/e

K=18/8

K=2.25lb/in

Work done by spring is given by

W=1/2Fe

Or W=1/2ke²

Then,

Work done in stretching the spring to 14in

W=1/2ke²

W=0.5×2.25×14²

W=220.5lbin

1 Inch-pounds Force to Joules = 0.113J

Then, to joules

W=0.133×220.5

W=29.33J

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