Tutorial Exercise A force of 7 lb ls required to hold a spring stretched 3 In. beyond Its natural length. How much work W is done in stretching It from its natural length to 9 in. beyond Its natural length? Step 1 According to Hooke's Law, the force required to maintain a spring stretched x units beyond Its natural length is proportional to x. This means that x) = kx, for some constant k. Since our spring is stretched 3 in., which is equal to I E: 1/4 R, we have 7 = %k. 50 k = l '1 E: 28 lb/ft. Now, we can say that the work done in stretching the spring 9 in. = 9/12 F: beyond its natural length is given by the following. 9/12 W=/ J 7 X dx 0 Submit Skip (you cannot come back)