Question

A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is filled with 36 lb of water and is pulled up at a rate of 2 ft/s, but water leaks out of a hole in the bucket at a rate of 0.2 lb/s. Find the work done in pulling the bucket to the top of the well. Show how to approximate the required work by a Riemann sum. (Let x be the height in feet above the bottom of the well. Enter xi* as xi.)

Answer 1

Given: wt. of the buclect = 5 lb Depth of well .=80 ft water in the buclect = 36 lb Rate of pulling aft/ see . Rate of healeage 0.2 lb/ see. Here to find Total work done in pulling the buclect with water, we should find o work done in pulling empty bucked © work done is lifting water Total work done, W = Woucket + Whater - C ost D K --- -- L4180ft - - (1 weight of buckeet = 5 lb. "bulleet = s 5 dx = 5[x] -5[80-o] 80 w bucket -1400

(2) For water: Rate of pulling = 2ft/ see i in second, water moves for 2 feet is water moves to 80 ft = 80 = 40 see water moves to the top in 40 seconds. But water reaks out oo2 lb / see. ... in 40 seconds, water leaks out (662x40) = 80 lb . Amount of water leaked out = 846 :: Amount of weiles remained = 36-8-2815 from the figure when x=0, weight = 2845 when r = 80, weight = 36 lb. 3. Slope = 26-28 cm
k
Linear equation of a line is y = matc -) weight = ma + weight = 16x +C When weight = 28 lb, we have a=0. ... 28 = 1 (0)+Ć 3) (228. .. weight - L X+28 :: Weveltek 5(+28) dx = $wdx+28 fo ( 39 ) +28( = (3032 +28(80) – 0 20 320 + 2040 = 2360 : wwater = 2360 Total work done = Wheelest t weetee
Hence, Total Work done = 400+2360 = 2760

Hence work done = 2760 ft-lb