The work required to stretch a spring x inches beyond its natural length is proportional to the square of x. Since it requires 6 ft-lb to stretch the spring one foot, it requires 6 × (3/4)2 = 27/8 ft-lb to stretch the spring 9 inches, or 3/4 of a foot.
We can verify the claim about the work required to stretch a spring being proportional to the square of the distance it is stretched by noting that the force the spring exerts is proportional to the distance it is stretched. Since W = ∫0x F ds and F = ks,where k is the spring constant, we have W = (kx2)/2. We can see that the spring constant for this particular spring is 12 lb/ft by substituting x = 1 and W = 6 in this equation.
