Work: Lifting A Load 100 ft Up From Below The Surface Of The Earth
Strings (SiPjAjk) = S7P3A31 Base Sequence = 12735 String Sequence = 12735 - 3 - 31
A cable weighing 2 lbs/ft is used to lift a 300 lbs load from the bottom of a well 100 feet deep.
Calculate the work done to raise the load to the surface.
(a) S7P3A31 (Force - Pull).
Pj Problem of interest is of type force. Work problems are generally of type force because force is the doer of work. In the case of white-collar work, intellectual force actuates materiality.
Consider the above diagram (fig. CW.1):
Let W be the work done to raise the load x feet.
Let k be the work done in raising the load h feet.
k = [300 + 2(100 - xavg)]h -------(1). xavg is the representative average of the partitions of the distance BC.
So, the rate of change of k with respect to h is:
k/h = [300 + 2(100 - xavg)] = 500 -2xavg
limit (k/h) as h → 0 = dW/dx (instantaneous change of W with respect to x)
So, limit (k/h) as h → 0 = 500 -2x. Since xavg → x.
So, dW/dx = 500 -2x.
So, W = 500x - x2 + C---------(2). Where C is a constant.
When x = 0, W = 0. So C = 0.
So, W = 500x - x2.
So, for x = 100,
W = 500(100) - (100)(100) = 40,000 ft-lb.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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