/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A swimming pool is \(20 \mathrm{... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 32

A swimming pool is \(20 \mathrm{m}\) long and \(10 \mathrm{m}\) wide, with a bottom that slopes uniformly from a depth of \(1 \mathrm{m}\) at one end to a depth of \(2 \mathrm{m}\) at the other end (see figure). Assuming the pool is full, how much work is required to pump the water to a level \(0.2 \mathrm{m}\) above the top of the pool?

Short Answer

Expert verified
The work required to pump the water out of the swimming pool is approximately \(4.74 \times 10^6\) Joules.

Step by step solution

01

Calculate the volume of water in the pool.

The pool has a trapezoidal shape in terms of depth. To find the volume, we can split the pool into a rectangular prism and a triangular prism: Rectangular volume: \(V_1=A_1l = 1 \mathrm{m} \times 10 \mathrm{m} \times 20 \mathrm{m} = 200 \mathrm{m^3}\) Triangular volume: \(V_2=\frac{1}{2}A_2l=\frac{1}{2}(1 \mathrm{m} \times 10 \mathrm{m}) \times 20 \mathrm{m} = 100 \mathrm{m^3} \) Total volume: \(V = V_1 + V_2 = 200 \mathrm{m^3} + 100 \mathrm{m^3} = 300 \mathrm{m^3} \)
02

Calculate the depth of water at position x.

We will use the distance x from the shallow end (depth of 1 m) to compute the depth of the water at that position: \(D(x) = 1\mathrm{m} + \frac{(2 \mathrm{m}-1 \mathrm{m})}{20 \mathrm{m}}x\) Simplifying the expression: \(D(x) = 1\mathrm{m} + \frac{1\mathrm{m}}{20 \mathrm{m}}x\)
03

Write the integral for work.

The work done on a small slice of water with thickness dx and width 10 m at position x can be expressed as: \(dW = \rho g A(x) D(x) (D(x) + 0.2 \mathrm{m}) dx\) where \(\rho\) is the density of water (\(1000 \mathrm{kg/m^3}\)), g is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)), and \(A(x)\) is the area of the slice (\(10 \mathrm{m} \times dx\)). Let's substitute the values and simplify: \(dW = 1000 \mathrm{kg/m^3} \cdot 9.81 \mathrm{m/s^2} \cdot 10 \mathrm{m} \cdot (1\mathrm{m} + \frac{1\mathrm{m}}{20 \mathrm{m}}x) \cdot (1\mathrm{m} + \frac{1\mathrm{m}}{20 \mathrm{m}}x + 0.2 \mathrm{m}) dx\)
04

Integrate to find the total work.

Integrate from 0 to 20 m to find the total work: \(W = \int_0^{20\mathrm{m}} 1000 \mathrm{kg/m^3} \cdot 9.81 \mathrm{m/s^2} \cdot 10 \mathrm{m} \cdot (1\mathrm{m} + \frac{1\mathrm{m}}{20 \mathrm{m}}x) \cdot (1\mathrm{m} + \frac{1\mathrm{m}}{20 \mathrm{m}}x + 0.2 \mathrm{m}) dx\) Simplifying and evaluating the integral: \(W = 98100 \int_0^{20\mathrm{m}} (1+\frac{x}{20})(1+\frac{x}{20}+0.2) dx\) \(W = 98100 \int_0^{20\mathrm{m}} (1+\frac{x}{20})(1.2+\frac{x}{20}) dx\) \(W = 98100 \int_0^{20\mathrm{m}} (1.2+\frac{41x}{400}+\frac{x^2}{400}) dx\) \(W = 98100 \left[\frac{1.2x+\frac{41}{800}x^2+\frac{1}{1200}x^3\right]_0^{20\mathrm{m}}\) Substitute the limit: \(W= 98100 (\frac{1.2\cdot20+ \frac{41}{800}\cdot20^2 + \frac{1}{1200}\cdot20^3} - 0) \) After solving the calculation: \(W \approx 4.74\times10^6 \mathrm{J}\) The work required to pump the water to a level 0.2 m above the top of the pool is about \(4.74\times10^6\) Joules.

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