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


Suppose
to represent the
subinterval, then the work done can be estimated as:


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



Therefore;
dW = f(x) dx
where x ranges from 0 to 500

![W =\bigg [750x + 4x^2 \bigg ]^{500}_{0}](https://tex.z-dn.net/?f=W%20%3D%5Cbigg%20%20%5B750x%20%2B%204x%5E2%20%5Cbigg%20%5D%5E%7B500%7D_%7B0%7D)
![W =\bigg [750(500) + 4(500)^2-0 \bigg ]](https://tex.z-dn.net/?f=W%20%3D%5Cbigg%20%20%5B750%28500%29%20%2B%204%28500%29%5E2-0%20%5Cbigg%20%5D)
W = 375000 + 1000000
W = 1375000
Thus; the total work done W = 13.8 × 10⁵ lb/ft