Aforce of 11 lb is required to hold a spring stretched 3 in. beyond its natural length. how much work w is done in stretching it from its natural length to 6 in. beyond its natural length?

5.5 lb-ft

Explanation:

k = Spring constant

x = Stretched length = 3 in = 3/12 = 1/4 feet

F = Force = 11 lb

So,

Work done when x = 6 in = 6/12 = 0.5 feet

Work done is 5.5 lb-ft

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