A bucket that weighs 4 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 38 lb of water and is pulled up at a rate of 1.5 ft/s, but water leaks out of a hole in the bucket at a rate of 0.15 lb/s. Find the work done W in pulling the bucket to the top of the well.

Evaluate the integral.
W = ft-lb

The total work done in in pulling the bucket to the top of the well is 3040ft-lb

Step-by-step explanation:

Given

Weight of bucket = 4lb

Depth of well = 80ft

So, Work done = Force * Distance

So, Work done to lift the bucket = Weight of buckets * Depth of the well

W1 = 4 * 80

W1 = 320ft-lb

Calculating the work done in pulling the bucket to the top of the well

At time t, the bucket is 1.5 ft above its original depth of 80ft

Also at time t, it now holds (38 - 0.15t) lbs of water.

Mathematically, we have

xi = 1.5t

xi = 3/2t

t = ⅔xi

In terms of distance, the bucket now holds

38 - 0.15 * ⅔xi lb off water when it is xi ft above its original depth of 80ft

Moving this water along a distance of ∆x requires a work done of

38 - 0.15(⅔xi) ∆x

= (38 - 0.3/3xi) ∆x

= (38 - 1/10xi) ∆x

= (38 - xi/10)∆x

Integrating the above we have

W2 = Integral of (38 - x/10) dx. Upper bound= 80 and lower bound = 0

(i.e from the depth (80ft) to the top (0ft))

Integrating, we have.

W2 = 38x - x²/20 (800)

=>

W2 = [38(80) - 80²/20] - [38(0) - 0²/20)]

W2 = [3040 - 320] - (0)

W2 = 2720 - 0

W2 = 2720ft-lb

Total work done in in pulling the bucket to the top of the well.

Total Work = W1 + W2

Total Work = 320ft-lb + 2720ft-lb

Total Work = 3040ft-lb

1/2*qr*rs

step-by-step explanation: