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
[tex]\Deltax = \dfrac{b -a }{n}[/tex]
[tex]\Deltax = \dfrac{500- 0 }{n}[/tex]
Suppose [tex][x_{i-1},x_i ][/tex] to represent the [tex]i^{th}[/tex] subinterval, then the work done can be estimated as:
[tex]W_i = f(x_i) \Delta x[/tex]
[tex]W_i =(750 + 8x_i) \Delta x[/tex]
Therefore; the total work done in between all the n subintervals is:
[tex]W = \sum \limits ^n_{i=1} W_1[/tex]
[tex]W = \sum \limits ^n_{i=1} f(x_i) \Delta x[/tex]
[tex]W = \sum \limits ^n_{i=1}(750 +8x_I)\Delta x[/tex]
Therefore;
dW = f(x) dx
[tex]\int \ dW = \int \ f(x) \ dx[/tex] where x ranges from 0 to 500
[tex]W = \int ^{500}_{0} \ 750 + 8x \ dx[/tex]
[tex]W =\bigg [750x + 4x^2 \bigg ]^{500}_{0}[/tex]
[tex]W =\bigg [750(500) + 4(500)^2-0 \bigg ][/tex]
W = 375000 + 1000000
W = 1375000
Thus; the total work done W = 13.8 × 10⁵ lb/ft
