Work require to pull bucket to the top of well?

Four times I tried to answer this question, but question keeps getting deleted. First few times, I thought perhaps poster deleted question because he wanted to repost in hopes of getting an answer, but last time question had only been posted for about 20 minutes.

Question:

A leaky bucket that weighs 6 lb and a rope of neglible weight are used to draw water from a well that is 373 ft deep. The bucket is filled with 35 lb of water and is pulled up at a rate of 5 ft/sec, but water leaks out of the bucket at a rate of 0.5 lb/sec. Find the work done in pulling the bucket to the top of the well (in ft-lb).

Additional Details (included last time question posted):

This is what I've done:

Force at time t: 6 lbs + 35 lbs - 0.5 lbs/sec * t sec = (41 - 0.5t) lbs

F(t) = 41 - 0.5t

Now we will express F in terms of distance x:

v = x/t = 5

t = x/5

F(x) = 41 - 0.5t x/5

F(x) = 41 - 0.1 x

Integrating from 0 to 373, I got:

W = ∫ (44 - 0.1x) dx from 0 to 373

W = 8336.55 ft-lbs

It says its not the correct answer. What am I doing wrong?

_______________________________

My answer:

What you are doing wrong is forgetting to take into account that eventually, bucket becomes empty

Here's my answer, which I tried to post before:

Let t = time (in sec)

Let h = height of bucket (in feet) from bottom of well -----> h varies from 0 to 373

Bucket is pulled up at a rate of 5 ft/sec -----> dh/dt = 5 -----> h = 5t -----> t = h/5

Let w = weight of bucket and water (in lbs), w(0) = 6 + 35 = 41

Water leaks out of bucket at a rate of 0.5 lb/sec -----> dw/dt = -0.5

w = 41 - 0.5t

w = 41 - 0.5 (h/5)

w = 41 - 0.1 h

Since water leaks out at a rate of 0.1 lb/ft, and initial amount of water = 35lb,

it will take 35lbs/(0.1 lb/ft) = 350 ft for bucket to be completely empty

So we have piecewise function:

w(h) = 41 - 0.1 h . . . . . 0 ≤ h ≤ 350 <--------- this matches your equation

w(h) = 6 . . . . . . . . . . . . 350 < h ≤ 373 <----- weight of empty bucket only

Work required to lift bucket and water weighing w(h) lbs a distance of ∆h feet

W = w(h) ∆h ft-lb

Integrating from h = 0 to h = 373, we get:

W = ∫₀³⁷³ w(h) dh

W = ∫₀³⁵⁰ (41 - 0.1 h) dh + ∫₃₅₀³⁷³ (6) dh

W = (41h - 0.05h²) |₀³⁵⁰ + 6h |₃₅₀³⁷³

W = (41 * (350 - 0) - 0.05 * (350² - 0)) + 6 (373 - 350)

W = 14350 - 6125 + 138

W = 8,363 ft-lb

_______________________________

I guess I'll just send email to person who posted question, with a link to this question and answer. I'm done trying to answer question, but didn't want my answer to go to waste.

Mαthmφm