Question

A bucket that weighs 4lband a rope of negligible weight are used todraw water from a well that is 80ftdeep. The bucket is filled with 40lbof waterand is pulled up at a rate of 2ft/s, but water leaks out of a hole in the bucketat a rate of 0.2lb/s. Find the work done in pulling the bucket to the top ofthe well.

Answer 1

Workdone = 3200 lb.ft

Explanation:

We are told that the bucket is filled with 40 lb of water but water leaks out of a hole in the bucket at a rate of 0.2lb/s

Thus,

Weight of water at any given time (t) would be;

w(t) = 40 - 0.2t - - - - (1)

We are told the bucket is pulled up at a rate of 2ft/s.

Thus, height at time (t); y = 0 + 2t = 2t

Since y = 2t,

Then,t = y/2

Put y/2 for t in eq 1

Thus; w(y) = 40 - 0.2(y/2)

w(y) = 40 - 0.1y

Now, at y = 80 ft, we have;

w(80) = 40 - 0.1(80)

w(80) = 40 - 8 = 32 lb

Since 32 lbs are left, it means there is always water in the bucket.

Thus, work done is;

W = 80,0[∫(Total weight).dy]

W = 80,0[∫[(weight of rope) + (weight of bucket) + (weight of water)]dy]

W = 80,0[∫[0 + 4 + 40 - 0.1y]dy]

Integrating, we have;

W = [44y - y²/20] at boundary of 80 and 0

So,

W = [44(80) - 80²/20] - [0 - 0²/20]

W = 3200 lb.ft