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

Answer 1

The general integral setup for this type of work problem is:

Work = ∫k*y dy from 0 to b.

Where k is the linear density of the cable (in lb/ft in this case), y is the distance from the top of the cable, and b is the total distance the weight at the bottom of the cable must travel.

With this setup, the top of cable must be labeled y = 0, and the bottom of the cable (attached to the coal weight) is labeled y = b. In this case, k = 2 ft/lb and b = 500 ft.

The 800 lb coal weight is an additional piece of information we have not yet accounted for.

Since lbs are a unit of force, the force to lift the coal is constantly equal to 800 lb, so the 800 lb weight can be added to the integral and the work to lift just the coal is the force * distance. The final integral is:

Work = ∫(2y + 800) dy from 0 to 500.

Integrating:

Work = y² + 800y | 0 to 500

= (500)² + 800(500)

= 650,000 ft-lb