A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 50 ft deep. The bucket is filled with 38 lb of water and is pulled up at a rate of 1.5 ft/s, but water leaks out of a hole in the bucket at a rate of 0.15 lb/s. Find the work done in pulling the bucket to the top of the well. Show how to approximate the required work by a Riemann sum. (Let x be the height in feet above the bottom of the well. Enter xi* as xi.)
Answers
Answer:
Part A
The work done in pulling the bucket to the top of the well is 1,775 ft.·lb
Part B
By Reimann approximation, the work done is 2,130 ft.·lb
Step-by-step explanation:
Part A
The parameters of the filled bucket of water being pulled upwards are;
The mass of the bucket = 5 lb
The mass of the rope = Negligible
The mass of water that fills the bucket = 38 lb
The rate at which the bucket is pulled up = 1.5 ft./s
The rate at which the bucket leaks, l = 0.15 lb/s
The height in feet above the bottom of the well = x
Therefore, we have;
Height of the bucket, x = 1.5·t
The weight of the bucket after a given time, t, W = 38 - 0.15·t
The time it takes to pull the bucket to the top of the well, t = d/v = 50/1.5 = 100/3 s. = [tex]33.\overline3[/tex] s
We have;
[tex]W = \int\limits^{t = 33.\overline 3}_{t = 0} {F} \, dx[/tex]
x = 1.5·t, F = Weight of the bucket, W = 38 - 0.15·t
[tex]W = \int\limits^{t = 33.\overline 3}_{t = 0} {(38-0.15\cdot t)} \, \cdot 1.5\cdot dt = \int\limits^{t = 33.\overline 3}_{t = 0} {(57-0.225\cdot t)} \, dt[/tex]
[tex]W = [57\cdot t - 0.225 \cdot t^2]_0^{33.\overline 3}[/tex]
Therefore, we have;
W = (57 × [tex]33.\overline3[/tex] - (0.225/2) × [tex]33.\overline3[/tex]²) - 0 = 1,775 ft.·lb
The work done in pulling the bucket to the top of the well, W = 1,775 ft.·lb
Part B
By summing the areas of the rectangular areas on the graph, we have
W = 10 × 38 + 10 × (38 - 0.1×10) + 10 × (38 - 0.1 × 20) + 10 × (38 - 0.1 × 30) + 10 × (38 - 0.1 × 40) + 10 × (38 - 0.1 × 50) = 2,130
Therefore, by left Reimann approximation, the work done, W = 2,130 ft.·lb
(More precisely, we have;
The weight at the bottom, F₁ = 38 lb.
The weight at the top of the well, F₂ = 38 - 0.15 × 100/3 = 33
∴ F₂ = 33 lb.
The average weight, [tex]F_{average}[/tex] = (F₁ + F₂)/2 = (38 lb. + 33 lb.)/2 = 35.5 lb.
Therefore, the work done, [tex]F_{average}[/tex] × d = 35.5 lb × 50 ft. = 1,775 ft.lb
The work done, W = 1,775 ft.lb)
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Answers
Answer:
8
Step-by-step explanation:
6÷[tex]\frac{3}{4}[/tex]
To make it easier, let's write 6 as a fraction.
[tex]6=\frac{6}{1}[/tex]
Now we get:
[tex]\frac{6}{1}[/tex]÷[tex]\frac{3}{4}[/tex]
Flip the numbers of [tex]\frac{3}{4}[/tex] and multiply instead of dividing and we get:
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Answers
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Step-by-step explanation:
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The value of y is 55°.
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The slope of the linear function y= 5/4x +1/2 is changed to 5/8 where y= 5/8x + 1/2. Identify the transformation required to produce the new slope, state the equation of the transformed function, ande explainw hat the graph of the tranfsformed line look like.
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Answers
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.jdnxixsykxkxmxmxmmxmmsmsmsmsmxmx
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