A force of 35 N is required to hold a spring that has been stretched from its natural EXAMPLE 3 length of 5 cm to a length of 10 cm. How much work is done in stretching the spring from 10 cm to 14 cm? According to Hooke's Law, the force required to hold the spring stretched x meters SOLUTION beyond its natural length is f(x) = kx. When the spring is stretched from 5 cm to 10 cm, the m. This means that f(0.05) = amount stretched is 5 cm = so 35 0.05k = 35 0.05 Thus f(x) 700x and the work done in stre ching the spring from 10 cm to 14 cm is ]0.09 r0.09 700 x2 2 Jo.05 700х dx 3D 350( ) - co.05)2 : J. = 350

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