Question

A cable that weighs 2 lbyft is used to lift 800 lb of coal up a mine shaft 500 ft deep. Find the work done.

Answer 1

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:

$W_{coal}$ = Weight of coal * distance

$W_{coal}$ = 800 * 500

$W_{coal}$ = 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:

$W_{shaft}$ = 2Δx * x

$W_{shaft}$ = 2xΔx

=> $W_{shaft}$ = $\int\limits^{500}_{0} {2x} \, dx$

$W_{shaft}$ = $\frac{2x^{2}}{2} \left \{ {{500} \atop {0}} \right.$

$W_{shaft}$ = $500^{2} - 0^{2}$

$W_{shaft}$ = 250000 ftlb

Therefore, total work done is:

$W_{total} = 250000 + 400000$

$W_{total} = 650000 ftlb$