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a force of 4lbs is required to hold a spring stretched 0.6 ft beyond its natural length. how much work (ft-lb) is done stretching the spring from its natural length to 1.1 ft beyond its natural length?
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do you know Hook's Law?
got it!:) 4.0315
Force = K * X where X is distance and K is the spring constant you can use F= 4 lbs when X= 0.6 ft to find K then find F when X = 1.1 work done will be F * X or F*1.1 or K*1.1^2
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