Answer:
W = 650000 ftlb
Step-by-step explanation:
Parameters given:
Weight of the cable = 2 lb/ft
Weight of the coal = 800 lb
Depth of the shaft, x = 500 ft
The total work done in this system is given by the sum of the work done on the coal and the work done on the shaft.
WORK DONE ON THE COAL
The work done on the coal is given as:
[tex]W_{coal}[/tex] = Weight of coal * distance
[tex]W_{coal}[/tex] = 800 * 500
[tex]W_{coal}[/tex] = 400000 ftlb
WORK DONE ON THE SHAFT
If we partition the cable into several segments of depth Δx, we can find the work done on the cable.
The work done on the shaft is given as:
[tex]W_{shaft}[/tex] = 2Δx * x
[tex]W_{shaft}[/tex] = 2xΔx
=> [tex]W_{shaft}[/tex] = [tex]\int\limits^{500}_{0} {2x} \, dx[/tex]
[tex]W_{shaft}[/tex] = [tex]\frac{2x^{2}}{2} \left \{ {{500} \atop {0}} \right.[/tex]
[tex]W_{shaft}[/tex] = [tex]500^{2} - 0^{2}[/tex]
[tex]W_{shaft}[/tex] = 250000 ftlb
Therefore, total work done is:
[tex]W_{total} = 250000 + 400000[/tex]
[tex]W_{total} = 650000 ftlb[/tex]
