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Answer 1 · 2018-05-18T14:18:35

See below

[Be careful about units. I have pretty much never used the old Imperial System before.]

The net upward force on the bucket, $F$ $F$ , is:

$F = T - (4 + m) g$ $F = T - ( 4 + m)g$

Newton's Second Law:

$F = \frac{d}{d t} (m v) = . m v + m . v = T - (4 + m) g$ $F = (d)/(dt) ( mv) = dot m v + m dot v = T - ( 4 + m)g$

The bucket travels at constant velocity $v = 2$ $v = 2$ , so then $. v = 0$ $dot v = 0$ :

Because the bucket is leaking at constant rate:

$(dm)/(dt) = - alpha, quad alpha > 0$

It also travels at constant velocity:

$v = (ds)/(dt) = 2$ , where $s$ is displacement in upward direction

Thus:

$(dm)/(ds) = (dm)/(dt) (dt)/(ds) = - alpha /2$ , which is also constant

So

$(dm)/(ds) = (Delta m)/(Delta s) = (30 - 50)/100 = -0.2 text( lb/ft)$

$implies m(s) = 50 - 0.2 s$

And

$dot m = - 0 .4 text( lb/sec)$

$= -0.2 g \ s + 54 g - 0.8$

The work done by $T$ in lifting the bucket is:

$= int_0^(100) -0.2 g s + 54 g - 0.8 \ ds$

$= ( -0.1 g\ s^2 + (54 g - 0.8)\ s )_0^(100) = 140720 text( lb)^2 text(ft)^2 text(sec)^(-2)$

Using $g = 32 \ ft* sec^(-2)$

Reality check:

Assuming no leak, it would be $54 * 32 * 100 = 172,000$ so it is correct order of magnitude and lower.
The bucket still retains $60 "%"$ of the water so you would expect it to be closer to the ideal no-leak position