Question

I have a Work word problem I need help completing?

Answer 1

See below

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

Explanation:

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

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

• `T` is the tension in the rope

• `m = m(t)` is the mass of water

Newton's Second Law:

`F = (d)/(dt) ( mv) = dot m v + m dot v = T - ( 4 + m)g`

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

• `implies 2 dot m = T - ( 4 + m)g`

• `implies T = 2 dot m + ( 4 + m)g`

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)`

• `implies T(s) = - 0.8 + ( 4 + (50 - 0.2 s))g`

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

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

• `W = int_0^(100) T(s) \ ds`

`= 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