Answer:
work = 1,830,000 ft·lb
Explanation:
You want the work done to lift 650 lb of coal 600 ft up a mine shaft using a cable that weighs 8 lb/ft.
Force
For some distance x from the bottom of the mine, the weight of the cable is ...
8(600 -x) . . . . pounds
The total weight being lifted is ...
f(x) = 650 +8(600 -x) = 5450 -8x
Work
The incremental work done to lift the weight ∆x feet is ...
∆w = force × ∆x
∆w = (5450 -8x)∆x
We can use a sum for different values of x to approximate the work. For example, the work to lift the weight the first 50 ft can be approximated by ...
∆w ≈ (5450 -8·0 lb)(50 ft) = 272,500 ft·lb
If we use the force at the end of that 50 ft interval instead, the work is approximately ...
∆w ≈ (5450 -8·50 lb)(50 ft) = 252,500 ft·lb
Sum
We can see that the first estimate is higher than the actual amount of work, because the force used is the maximum force over the interval. The second is lower than the actual because we used the minimum of the force over the interval. We expect the actual work to be close to the average of these values.
The attached spreadsheet shows the sums of forces in each of the 50 ft intervals. The "left sum" is the sum of forces at the beginning of each interval. The "right sum" is the sum of forces at the end of each interval. The "estimate" is the average of these sums, multiplied by the interval width of 50 ft.
The required work is approximated by 1,830,000 ft·lb.
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Additional comment
The actual work done is the integral of the force function over the distance. Since the force function is linear, the approximation of the area under the force curve using trapezoids (as we have done) gives the exact integral. It is the same as using the midpoint value of the force in each interval.
Because the curve is linear, the area can be approximated by the average force over the whole distance, multiplied by the whole distance:
(5450 +650)/2 × 600 = 1,830,000 . . . . ft·lb
Another way to look at this is from consideration of the separate masses. The work to raise the coal is 650·600 = 390,000 ft·lb. The work to raise the cable is 4800·300 = 1,440,000 ft·lb. Then the total work is ...
390,000 +1,440,000 = 1,830,000 . . . ft·lb
(The work raising the cable is the work required to raise its center of mass.)