Chapter 6.6, Problem 13E

To determine

To Calculate: The work done to lift a load of coal from a mine shaft using a heavy cable.

Answer to Problem 13E

Explanation of Solution

Given Information: The cable weighs $2 lbperfeet$ and the coal weighs $800lb$ . The depth of the shaft is $500ft$ .

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.

Calculate the work done on the cable and the work done on the load of coal separately and then add them

Work done on shaft:

The weight of an infinitesimally short element of the cable with length $dx$ and at a depthof $x$ from the ground is

  $dm=weightperunitlength\times dx=2\cdot dx$

Work required to pull this infinitesimally small element to the ground is

  $dw=F\cdot d=2\cdot dx\cdot x=2xdx$

The work required to pull the entire cable to the top of the building is the integral of $dw$ from $x=0$ for the ground level to $x=500ft$ for the deepest point of the cable.. Therefore

  $W_{cable}={{\displaystyle \int }}_0^{500}dw={{\displaystyle \int }}_0^{500}2xdx$

  $={x^2]}_0^{500}$

  $=\left({500}^2-0^2\right)$

  $=\underset{\_}{250,000}$

Work done on coal: Multiply the weight of the coal (force) with the depth of the coal (displacement) to get the work done on the coal to write

  $W_{coal}=weigh{t}_{coal}\times d_{coal}=800\times 500=\underset{\_}{400,000}$

So, the total work done is

  $W=W_{cable}+W_{coal}=250,000+400,000=650,000lb\cdot ft$