/*! 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] | 91Ó°ÊÓ

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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
Question: Calculate the work required to pump the water to a level 0.2 m above the top of the pool. Solution: We have already calculated the Initial and final potential energies for rectangular and triangular parts of the pool. Initial Potential Energy: PE(rectangular part) = 200m³ × 1000 kg/m³ × 9.8 m/s² × 0.5m = 980,000 J PE(triangular part) = 100m³ × 1000 kg/m³ × 9.8 m/s² × 1.5m = 1,470,000 J Initial Potential Energy = 980,000 J + 1,470,000 J = 2,450,000 J Final Potential Energy: PE(rectangular part) = 200m³ × 1000 kg/m³ × 9.8 m/s² × 0.7m = 1,372,000 J PE(triangular part) = 100m³ × 1000 kg/m³ × 9.8 m/s² × 1.7m = 1,666,000 J Final Potential Energy = 1,372,000 J + 1,666,000 J = 3,038,000 J Now, we can find the work done using the Work-Energy theorem. Work = Final Potential Energy - Initial Potential Energy Work = 3,038,000 J - 2,450,000 J Work = 588,000 J The work required to pump the water to a level 0.2 m above the top of the pool is 588,000 J (joules).

Step by step solution

01

Find the volume of the pool

Since the pool's bottom slopes uniformly, we can treat it as a triangular prism. We first need to separate the pool into two parts: a rectangular prism with dimensions 20 m x 10 m x 1 m, and a triangular prism with base 10 m, height 20 m, and a thickness of 1 m. For the rectangular part, the volume is given by: Volume = Length x Width x Height = 20m × 10m × 1m = 200 m³ For the triangular part, the volume is given by: Volume = (1/2) x Base x Height x Thickness = (1/2) × 20m × 10m × 1m = 100 m³ Now, sum the volume of both parts to get the total volume of the pool: Total Volume = 200m³ + 100m³ = 300 m³
02

Find the initial potential energy of the water

To find the initial potential energy, we need to find the center of mass of the pool and multiply by the volume, density, and gravitational acceleration. Let's consider the center of mass (COM) of these two parts separately. The COM of the rectangular part is at the center of the rectangle, where x=10m (half of the length along the sloping bottom). The COM of the triangular part is at x=30/3 = 10m. Now, we find the initial potential energy (PE) for both parts: PE(rectangular part) = Volume x Density x Gravitational acceleration x COM height = 200m³ × 1000 kg/m³ × 9.8 m/s² × 0.5m PE(triangular part) = Volume x Density x Gravitational acceleration x COM height = 100m³ × 1000 kg/m³ × 9.8 m/s² × 1.5m Now, sum the initial potential energies of both parts to get the initial potential energy of the water: Initial Potential Energy = PE(rectangular part) + PE(triangular part)
03

Find the final potential energy of the water

It's given that the water is pumped to a level 0.2 m above the top of the pool. So for both rectangular and triangular parts, we have to add an extra height of 0.2m for their COM height. Then, we can find the final potential energy. PE(rectangular part) = Volume x Density x Gravitational acceleration x COM height = 200m³ × 1000 kg/m³ × 9.8 m/s² × 0.7m PE(triangular part) = Volume x Density x Gravitational acceleration x COM height = 100m³ × 1000 kg/m³ × 9.8 m/s² × 1.7m Now, sum the final potential energies of both parts to get the final potential energy of the water: Final Potential Energy = PE(rectangular part) + PE(triangular part)
04

Find the work done

Now, we have the initial and final potential energies of the water. We can find the work done using the Work-Energy theorem. Work = Final Potential Energy - Initial Potential Energy Calculate the values, and the work done will be obtained.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangular Prism Volume
Understanding how to find the volume of a triangular prism is essential in fields such as geometry and physics. A triangular prism is a three-dimensional figure with two triangular bases and three rectangular sides, resembling a stretched-out triangle.

The volume of a triangular prism can be found using the formula: \[ \text{Volume} = \frac{1}{2} \times \text{Base} \times \text{Height} \times \text{Length} \] where the 'Base' and 'Height' refer to the triangle's base and height, respectively, and 'Length' corresponds to the distance between the triangular bases (also known as the 'prism height' or 'thickness').

For example, in the swimming pool problem described earlier, the volume of the triangular part of the pool was calculated by taking half of the product of the base width (10 m), the distance along the sloping bottom (20 m), and the height difference between the deep and shallow ends (1 m). Simplifying this provides us with a volume of 100 cubic meters for the triangular section of the pool.
Potential Energy Calculation
Potential energy (PE) is the stored energy in an object due to its position or state. In the context of pumping water from a pool, potential energy is related to the elevation of the water above a reference point, typically the ground.

The general formula for the gravitational potential energy for an object at height 'h' is: \[ PE = m \times g \times h \] where 'm' is the mass of the object, 'g' is the acceleration due to gravity (9.8 m/s² on Earth), and 'h' is the height above the reference point.

When dealing with liquids such as water, mass is often replaced with volume multiplied by density (since \( m = \text{density} \times \text{volume} \)). For the swimming pool exercise, to find the potential energy of the water, you would multiply the volume of water in each section of the pool by the water's density, gravitational acceleration, and the averaged height of the water's mass (its center of mass) above the reference point.
Work-Energy Theorem
The work-energy theorem is a fundamental principle of physics that relates the work done on an object to a change in its kinetic or potential energy. In simple terms, the theorem states that work done by forces on an object results in a change in the object's energy.

Mathematically, the work done 'W' is the difference in energy, and for potential energy in cases like pumping water it is: \[ W = \text{PE}_{\text{final}} - \text{PE}_{\text{initial}} \] In the scenario of pumping water out of a pool, we calculate the work done as the change in potential energy of the water from its initial state at the pool's bottom to its final state at a specified height above the pool. The step-by-step solution for the swimming pool shows how to calculate this work by finding the difference in potential energy before and after pumping the water.
Gravitational Acceleration
Gravitational acceleration is the acceleration caused by the gravitational pull of large bodies, like Earth. It is denoted by 'g' and is approximately \(9.8 \text{m/s}^2\) on the surface of our planet.

This acceleration is what imparts weight to objects and is a critical factor in calculating forces in the field of mechanics. For example, the force exerted by the water's weight in the pool, which acts downward due to gravity, is a product of the mass of the water and gravitational acceleration.

In energy terms, gravitational acceleration plays a vital role in determining the potential energy of an object at a certain height. In the context of the pool problem, 'g' is used in calculating the potential energy of the water, and hence the work required to pump it to a certain level above the pool.

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Most popular questions from this chapter

A typical human heart pumps \(70 \mathrm{mL}\) of blood with each stroke (stroke volume). Assuming a heart rate of 60 beats \(/ \min (1 \text { beat } / \mathrm{s}),\) a reasonable model for the outflow rate of the heart is \(V^{\prime}(t)=70(1+\sin 2 \pi t),\) where \(V(t)\) is the amount of blood (in milliliters) pumped over the interval \([0, t]\) \(V(0)=0,\) and \(t\) is measured in seconds. a. Graph the outflow rate function. b. Verify that the amount of blood pumped over a one-second interval is 70 mL. c. Find the function that gives the total blood pumped between \(t=0\) and a future time \(t>0\) d. What is the cardiac output over a period of 1 min? (Use calculus; then check your answer with algebra.)

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