# I have a Work word problem I need help completing?

## May 18, 2018

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:

$F = T - \left(4 + m\right) g$

• $T$ is the tension in the rope

• $m = m \left(t\right)$ is the mass of water

Newton's Second Law:

$F = \frac{d}{\mathrm{dt}} \left(m v\right) = \dot{m} v + m \dot{v} = T - \left(4 + m\right) g$

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

• $\implies 2 \dot{m} = T - \left(4 + m\right) g$

• $\implies T = 2 \dot{m} + \left(4 + m\right) g$

Because the bucket is leaking at constant rate:

$\frac{\mathrm{dm}}{\mathrm{dt}} = - \alpha , \quad \alpha > 0$

It also travels at constant velocity:

$v = \frac{\mathrm{ds}}{\mathrm{dt}} = 2$, where $s$ is displacement in upward direction

Thus:

$\frac{\mathrm{dm}}{\mathrm{ds}} = \frac{\mathrm{dm}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{ds}} = - \frac{\alpha}{2}$, which is also constant

So

$\frac{\mathrm{dm}}{\mathrm{ds}} = \frac{\Delta m}{\Delta s} = \frac{30 - 50}{100} = - 0.2 \textrm{l \frac{b}{f} t}$

$\implies m \left(s\right) = 50 - 0.2 s$

And

$\dot{m} = - 0 .4 \textrm{l \frac{b}{\sec}}$

• $\implies T \left(s\right) = - 0.8 + \left(4 + \left(50 - 0.2 s\right)\right) g$

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

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

• $W = {\int}_{0}^{100} T \left(s\right) \setminus \mathrm{ds}$

$= {\int}_{0}^{100} - 0.2 g s + 54 g - 0.8 \setminus \mathrm{ds}$

$= {\left(- 0.1 g \setminus {s}^{2} + \left(54 g - 0.8\right) \setminus s\right)}_{0}^{100} = 140720 {\textrm{l b}}^{2} {\textrm{f t}}^{2} {\textrm{\sec}}^{- 2}$

Using $g = 32 \setminus f t \cdot {\sec}^{- 2}$

Reality check:

• Assuming no leak, it would be $54 \cdot 32 \cdot 100 = 172 , 000$ so it is correct order of magnitude and lower.

• The bucket still retains $60 \text{%}$ of the water so you would expect it to be closer to the ideal no-leak position