A cable that weighs 8 lb/ft is used to lift 900 lb of coal up a mine shaft 650 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum.

2275000 lb.ft

Explanation:

Let work done on the cable be denoted by: W_ca

Let work done on the coal be denoted by: W_co

Now, dividing the cable into segments, let x represent the length from top of the mine shaft to the segment.

Meanwhile let δx be the length of the segment.

We are told the cable weighs 8 lb/ft. Thus;

Work done on one segment = 8 × δx × x = 8x•δx

Therefore, work done on cable is;

W_ca = ∫8x•δx between the boundaries of 0 and 650

Thus;

W_ca = 4x² between the boundaries of 0 and 650

W_ca = 4(650²) - 4(0²)

W_ca = 1,690,000 lb.ft

Workdone on the 900 lb of coal will be calculated as;

W_co = 900 × 650

W_co = 585000 lb.ft

Thus,

Total work done = W_ca + W_co

Total workdone = 1690000 + 585000

Total workdone = 1690000 + 585000

Total workdone = 2275000 lb.ft

kinetic

i'm so sorry if this is wrong i'm only in 6th grade xd

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6