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## A force of 12 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 n

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## Answers ( )

Answer:Work done in stretching the spring = 7.56 lb-ft.

Step-by-step explanation:Normal length of the spring = 8 in or ft

= ft

If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

= ft

Force applied to stretch the spring = 12 lb

By

Hook’s law,F = kx [where k is the spring constant and x = length by which the spring is stretched]

12 = k()

k = 18

Work done (W) to stretch the spring by ft will be

W =

=

=

= 9()²

=

7.56 lb-ft