Question

a cable that weighs 6 lb/ft is used to lift 650 lb of coal up a mine shaft 300 ft deep. find the work done. show how to approximate the required work by a riemann sum. (let x be the distance in feet below the top of the shaft. enter x_i^* as x_i.) lim_(n→∞) sum_(i = 1)^n ( )δx express the work as an integral. ∫_0^300 ( )dx evaluate the integral. ft - lb

Explanation:

Step1: Determine the work - force and distance relationship

The force required to lift a small section of the cable and the coal is the sum of the force to lift the coal and the force to lift the cable. The force to lift the coal is a constant $F_{coal}=650$ lb. The force to lift a small section of the cable of length $\Delta x$ at a depth $x$ is $F_{cable}=6\Delta x$. The work done in lifting a small section of the cable and coal through a distance $\Delta x$ is $\Delta W=(650 + 6x)\Delta x$.

Step2: Set up the Riemann sum

The Riemann sum for the work is $\sum_{i = 1}^{n}(650+6x_{i})\Delta x$, where $x_{i}$ is a sample - point in the $i$ - th sub - interval and $\Delta x=\frac{300}{n}$, and the limit as $n
ightarrow\infty$ gives the integral.

Step3: Express the work as an integral

The work $W$ is given by the definite integral $\int_{0}^{300}(650 + 6x)dx$.

Step4: Evaluate the integral

We use the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$ and $\int kdx=kx + C$ (where $k$ is a constant).
\[

$$\begin{align*} \int_{0}^{300}(650 + 6x)dx&=\int_{0}^{300}650dx+\int_{0}^{300}6xdx\\ &=650x\big|_{0}^{300}+6\times\frac{x^{2}}{2}\big|_{0}^{300}\\ &=650\times(300 - 0)+3\times(300^{2}-0^{2})\\ &=650\times300+3\times90000\\ &=195000+270000\\ &=465000 \end{align*}$$

\]

Answer:

$465000$ ft - lb