SCAPFYS. A.A. thisFrample 4Gheep frample 41A EQS-B cable is 100 In long and hangs verically from the top ef a tall building.(a) How thuch work is required to in the cable to the top of the building?(B) How much work is required to pull we enly is feet of the catie?Selvition(8)- there me don have a formula for the force function, bot we can whe an argument similar to the cree that led to the definition of mork.text place the arigin at the top of the bulding and the x-axis pointing downmard as in the following figure.We divide the cable into small parts with length x. If xi** is a point in the r th such interval, then all pcints in the interval are lifted by approximately the same amount, namely xi+. The cable weighs pounds per foot, so the weight of the /th part is 6 Jr . Thus the work done on thei th part, in. foot-pounds, is(6x)=xi=6xi2xforce distanceWe get the total work done by adding all these approximations and letting the number of parts become large (so x0).W=limnj=1n6xj2x=01006xdx=0100=(b) The work required to move the top 15 ft of cable to the top of the bulling is computed in the same manner as part (a).W1=0156xdx=3x2|015|=,r-10Every part of the lower 85 ft of cable moves the same distance, namiely 15 ft , so the work done isW2=limnj=1n([15,6x],[ distance , force ])=1510090dx=(A)ternatively, we can observe that the lower 85 ft of cable weighs 85-6=510 ib and moves uniformly 15 ft , so the work done is 510*15=,ft-ib.)??????(f-lb.)The total work done is w1+w2=675+??????=8,325 R-ID.Resourcesus