EXAMPLE 3 A force of 60 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How much work is done in stretching the spring from 35 cm to 40 cm? SOLUTION According to Hooke's Law, the force required to hold the spring stretched x meters beyond its natural length is x} 2 kx. When the spring is stretched from 20 cm to 35 cm, the amount stretched is 15 cm 2 m. This means that f(0.15) = , so U.15k=60 k=i= 0.15 Thus f(x) = 400X and the work done in stretching the spring from 35 cm to 40 cm is 0.2 2 0.2 _ x f 400x dx 400] 0 2 .15 015 = 200m: )2 (0.132] = J. W