Question

A cable that weighs 8 lb/ft is used to lift 750 lb of coal up a mine shaft 500 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 xi* as xi.)

Answer 1

Answer:

13.8 × 10⁵ lb/ft

Step-by-step explanation:

Suppose the distance(ft) below the top of the shaft is represented by x

The weight of the cable = 8 lb/ft

Weight of the coal to be lifted from mine = 750 lb

Recall that:

The work done by a force f to move an object through a distance x can be expressed as:

W = force (f) × displacement (x)

So, the force implies the total weight which should be lifted at any height x

f(x) = 750 + 8x

Using Riemann sum

where; the coal is lifted from x = 0 to x = 500 i.e. [a,b] = [0,500]

Dividing the interval into n subintervals

$\Deltax = \dfrac{b -a }{n}$

$\Deltax = \dfrac{500- 0 }{n}$

Suppose $[x_{i-1},x_i ]$ to represent the $i^{th}$ subinterval, then the work done can be estimated as:

$W_i = f(x_i) \Delta x$

$W_i =(750 + 8x_i) \Delta x$

Therefore; the total work done in between all the n subintervals is:

$W = \sum \limits ^n_{i=1} W_1$

$W = \sum \limits ^n_{i=1} f(x_i) \Delta x$

$W = \sum \limits ^n_{i=1}(750 +8x_I)\Delta x$

Therefore;

dW = f(x) dx

$\int \ dW = \int \ f(x) \ dx$ where x ranges from 0 to 500

$W = \int ^{500}_{0} \ 750 + 8x \ dx$

$W =\bigg [750x + 4x^2 \bigg ]^{500}_{0}$

$W =\bigg [750(500) + 4(500)^2-0 \bigg ]$

W = 375000 + 1000000

W = 1375000

Thus; the total work done W = 13.8 × 10⁵ lb/ft