Question

1. 0.26 / 1.07 points

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 $x_i$.)
$lim_{n
ightarrowinfty}sum_{i = 1}^{n}(quad)delta x$
express the work as an integral.
$int_{0}^{300}(quad)dx$
evaluate the integral.
$quad$ft - lb

Explanation:

Step1: Analyze work - force and distance

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 distance $x_i$ from the top of the shaft is $F_{cable - i}=6\Delta x$ lb. The work done in lifting the coal is $W_{coal}=650\times300$ ft - lb. For the cable, we use a Riemann sum. The work done in lifting a small section of the cable of length $\Delta x$ through a distance $x_i$ is $dW_i = 6x_i\Delta x$. The Riemann sum for the work done on the cable is $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(6x_i)\Delta x$.

Step2: Express work as an integral

The work done on the cable is $W_{cable}=\int_{0}^{300}6x dx$. The total work $W$ is the sum of the work done on the coal and the work done on the cable. So $W=\int_{0}^{300}(650 + 6x)dx$.

Step3: Evaluate the integral

We know that $\int_{0}^{300}(650+6x)dx=\int_{0}^{300}650dx+\int_{0}^{300}6x dx$.
For $\int_{0}^{300}650dx=650x\big|_{0}^{300}=650\times300 - 650\times0 = 195000$.
For $\int_{0}^{300}6x dx=6\times\frac{x^{2}}{2}\big|_{0}^{300}=3x^{2}\big|_{0}^{300}=3\times(300^{2}-0^{2})=3\times90000 = 270000$.
Then $W=195000 + 270000=465000$ ft - lb.

Answer:

$\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(6x_i)\Delta x$; $\int_{0}^{300}(650 + 6x)dx$; $465000$