Aforce 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 natural length to 11 in. beyond its natural length?

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

42

step-by-step explanation:

256/6=42