Example 4 Video Example-m A GOD-lb cable is 100 ft long and hangs vertically from the top of a tall building. (a) How much work is required to lift the cable to the top of the building? (b) How much werk is required to pull up only 10 feet of the cable? Solution (a) Here we don't have a formula for the force function, but we can use an argument similar to the one that led to the denition of work. Let's place the origin at the top of the building and the x-axis pointing downward as in the following gure. Q) We divide the cable into small parts with length Ax. Ifx: is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely x;. The cable weighs :l pounds per foot, so the weight of the i th part is 6Ax. Thus the work done on the ith part, in foot-pounds, is (6AX)' x.' = 6fox I force dlsta rice We get the total work clone by adding all these approximations and letting the number of parts become large (so Ax > 0). W II a: $3 [:1 HR E? 100 100 f WSi =i:w Cl 0 (b) The work required to move the top 10 ft of cable to the top of the building is computed in the same manner as part (a). 10 ll] w1=f 6xdx=3x2] =:]ft-lb 0 Cl Every part of the lower 90 ft of cable moves the same distance, namely 10 ft, so the work done is n 100 . 10 ' BAX W = llm = 60 dx = ft-lb. 2 - (distance force ) [10 E J: (Alternatively, we can observe that the lower 90 ft of cable weighs 90 - 2 = 540 lb and moves uniformly 10 ft, 50 the work done is 540 - 10 = l: ftlb.) The total work done is w1 + w2 = 300 + :J = 5,700 ftlb