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A force of 10 pounds stretches a spring 2 inches. Find the work done in stretching this spring 3 inches beyond its natural length

  • calculus/physics -

    Assuming an ideal spring (linear) and using Hooke's law, the force generated by the spring is:
    F = -kx (1)
    where x is the distance the spring is stretched from the rest position. So, substituting the info into (1) you get:
    -10 = -k(2)
    k = 5 lb/in
    work is the 1st integral of force with respect to x, which would be:
    W = (1/2)kx^2
    = (1/2)5(3)^2
    = 22.5 in-lb

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