Integral Calculus

A force of 4 pounds is required to hold a spring stretched 0.1 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?

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  1. work=INT force*dx from 0 to .6
    = int(k x dx)= 1/2 k x^2 over limits
    where k= 4lbs/.1f= 20 lbs/ft
    work= 1/2 k x^2 from zero to x=.6
    work= 1/2 (20lbs/ft)*.6^2 ft^2 - 0
    = 36 ft-lbs work

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    bobpursley

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