Chapter 6.6, Problem 14E

To determine

To Calculate: The work done in pulling up a leaking bucket of water from a well.

Answer to Problem 14E

The total work done in pulling the leaking bucket of water to the top of the well is $3200lb\cdot ft$ .

Explanation of Solution

Given Information:

• The weight of the bucket is $4lb$ . Weight of the rope is negligible.
• The depth of the well is $80ft$ .
• The quantity of water in the bucket at the bottom of the well is $40lb$ .
• The climbing rate of the bucket is $2ft/s$ .
• The leaking rate of water is $0.2lb/s$ .

Concept Used: The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

If the force is constant, then work done can be expressed as $F\cdot d$ , where $F$ is the force and $d$ is the displacement.

Calculation:

The time $t$ taken by the bucket to reach a height $h$ from the bottom is

  $t=\frac{height}{climbingrate}=\frac{h}{2}$

The amount of water leaked from the bucket in this time $t$ is

  $Q_{leaked}=leakingrate\times t=0.2\cdot \frac{h}{2}=0.1h$

So, the quantity of water remaining at height $h$ is

  $Q_{remaining}=Q_{initial}-Q_{leaked}=40-0.1h$

Therefore, the total weight of the bucket and water at a height $h$ is

  $totalweight=bucketweight+waterweight=4+\left(40-0.1h\right)=44-0.1h$

The work done to lift this total weight by an infinitesimally small height $dh$ when it is located at a height $h$ is

  $dW=totalweight\cdot distance=\left(44-0.1h\right)\cdot dh$

Then, the total work done $W$ is the integral of this infinitesimal work from a height $h=0$ to $h=80$ .

Therefore

  $W={{\displaystyle \int }}_0^{80}dW={{\displaystyle \int }}_0^{80}\left(44-0.1h\right)dh$

  $={44h-0.1\frac{h^2}{2}]}_0^{80}=\left(44\times 80-0.1\frac{{80}^2}{2}\right)-\left(0-0\right)$

  $=3200$

Therefore, the total work done in pulling this leaking bucket of water to the top of the well is $3200lb\cdot ft$ .