Mail - Her X A 6.4 Home X Course H X Course H X Course H X Smarthink x # integrate x Course H X Course H X Course H X Course H X Course H X Course H X + - C webassign.net/web/Student/Assignment-Responses/last?dep=27084138 R 2. [-/1 Points] DETAILS SCALCET9 6.4.AE.004. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Example 4 Video Example () A 600-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 work is required to pull up only 20 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 definition of work. Let's place the origin at the top of the building and the x-axis pointing downward as in the following figure. 100 We divide the cable into small parts with length Ax. If x, 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 pounds per foot, so the weight of the ith part is 6Ax. Thus the work done on the ith part, in foot-pounds, S (64x) . X; = 6x," Ax force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax -> 0). W = lim 6x 4x 100 6x dx = oor ft - lb (b) The work required to move the top 20 ft of cable to the top of the building is computed in the same manner as part (a). W1 = 2x dx = x2 =[ft - lb Every part of the lower 80 ft of cable moves the same distance, namely 20 ft, so the work done is W2 = lim 20 2Ax distance force 40 dx =[ ft - lb. (Alternatively, we can observe that the lower 80 ft of cable weighs 80 . 2 = 160 lb and moves uniformly 20 ft, so the work done is 160 . 20 = ft-lb.) The total work done is W1 + W2 = 400 + = 3,600 ft-lb. Need Help? Read It