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3. A work of 50 ft-lb stretches a spring 10 inches beyond its natural length. How much work is done to stretch 15 inches beyo

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2. 28,2 work done, AW = 1 KX 2 ² - 1 KX12, where, kx²= so AW = 100 (X2² - X, 2) AW = 50 ( (*21x.)-1) o W=so ((15/16) ? | sw =

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