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Chapter 6: Problem 7

Compute the work done against gravity by a pump that discharges 600 liters of fuel oil into a tank \(20 \mathrm{~m}\) above the pump's intake. One cubic centimeter of fuel oil has a mass of \(0.82 \mathrm{~g} .\) One liter is \(1000 \mathrm{~cm}^{3}\) The mass lifted is $$ (600 \text { liters })\left(1000 \frac{\mathrm{cm}^{3}}{\text { liter }}\right)\left(0.82 \frac{\mathrm{g}}{\mathrm{cm}^{3}}\right)=492000 \mathrm{~g}=492 \mathrm{~kg} $$ The lifting work is then Work \(=(m g)(h)=\left(492 \mathrm{~kg} \times 9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})=96 \mathrm{~kJ}\)

Short Answer

Expert verified
The work done is approximately 96 kJ.

Step by step solution

01

Understanding the Problem

We need to calculate the work done by a pump in lifting 600 liters of fuel oil to a height of 20 meters. The work done against gravity can be calculated using the formula for gravitational potential energy.
02

Convert Volume to Mass

First, convert the volume of fuel oil (600 liters) into mass. Since 1 liter equals 1000 cm鲁, we have 600,000 cm鲁. Knowing the density (mass per volume) of the oil is 0.82 g/cm鲁, the total mass in grams is calculated as:\[ \text{Mass} = 600000 \text{ cm}^3 \times 0.82 \frac{\text{g}}{\text{cm}^3} = 492000 \text{ g} \]Convert grams to kilograms:\[ 492,000 \text{ g} = 492 \text{ kg} \]
03

Calculating Work Done

Use the formula for gravitational work done: \( \text{Work} = mgh \).Where \( m = 492 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity), and \( h = 20 \text{ m} \).Calculate the work:\[ \text{Work} = 492 \text{ kg} \times 9.81 \text{ m/s}^2 \times 20 \text{ m} = 96477.6 \text{ J} \]Convert joules to kilojoules:\[ \text{Work} = 96477.6 \text{ J} \approx 96 \text{ kJ} \]

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. When we lift something against gravity, like the fuel oil in our problem, we're storing energy in it. This stored energy can be calculated using the formula:
  • \( ext{Potential Energy} = mgh \)
Where \(m\) is the mass, \(g\) is the acceleration due to gravity (9.81 m/s虏 on Earth), and \(h\) is the height above the reference point. This formula assumes that the gravitational field is uniform, which is a good approximation for small heights on Earth.
When the pump lifts the oil to the tank, it works against the pull of gravity. This work is essentially the energy stored in the lifted oil due to gravity.
Conversion of Units
In physics, converting units to consistent systems is crucial, as it allows for accurate calculations. For this exercise, began with the conversion of volume units to mass units. We started with the volume of fuel oil in liters and then converted this to cubic centimeters, since 1 liter is equal to 1000 cm鲁. This was essential since the given density was in grams per cubic centimeter (g/cm鲁).
After calculating the mass in grams, it was necessary to convert this mass into kilograms because the formula for gravitational potential energy uses kilograms as the standard unit for mass. In SI units, the conversion is simple: 1000 grams equals 1 kilogram. Utilizing consistent units is key for the straightforward application of formulas.
Density and Mass Calculation
Density is defined as mass per unit volume and is usually expressed in units like g/cm鲁 or kg/m鲁. In the problem, the density of the fuel oil is given as 0.82 g/cm鲁. To find the total mass of the oil, we use this density value with the calculated volume in cm鲁:
  • \( ext{Mass} = ext{Density} imes ext{Volume} \)
  • \( ext{Mass} = 0.82 rac{ ext{g}}{ ext{cm}^3} imes 600,000 ext{ cm}^3 \)
This gives us the total mass of fuel oil in grams, which then was converted into kilograms (492 kg). Calculating the mass correctly is a crucial step for ensuring that subsequent calculations for work and energy use accurate values.
Formula Application in Physics
Applying formulas correctly in physics problems involves ensuring all variables are in the appropriate units and plugged into the correct spots in the equation. In this context, the formula for calculating work done against gravity, \( ext{Work} = mgh \), is used after determining the correct values for mass, height, and acceleration due to gravity.
  • For mass \( m = 492 \text{ kg} \)
  • The height \( h = 20 \text{ m} \) above the pump's intake
  • Acceleration due to gravity \( g = 9.81 \text{ m/s}^2 \)
Substituting these values into the formula yields the result for work done in Joules. Converting to kilojoules by dividing by 1000 gives approximately 96 kJ.
Understanding how to correctly apply these physics formulas is essential for solving a wide range of real-world problems, exactly as was done here.

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Most popular questions from this chapter

How much work in total must a 200-lb man do climbing to the top of the 555 -ft-tall Washington Monument carrying a \(10.0\) -kg backpack? [Hint: \(1 \mathrm{lb}=4.448 \mathrm{~N}\).]

A \(1200-\mathrm{kg}\) car coasts from rest down a driveway that is inclined \(20^{\circ}\) to the horizontal and is \(15 \mathrm{~m}\) long. How fast is the car going at the end of the driveway if \((a)\) friction is negligible and \((b)\) a friction force of \(3000 \mathrm{~N}\) opposes the motion?

A 1300 -kg car is to accelerate from rest to a speed of \(30.0 \mathrm{~m} / \mathrm{s}\) in a time of \(12.0 \mathrm{~s}\) as it climbs a \(15.0^{\circ}\) hill. Assuming uniform acceleration, what minimum power is needed to accelerate the car in this way?

Imagine a 60.0-kg skier standing still on the top of a snow-covered hill \(150 \mathrm{~m}\) high. Neglecting any friction losses, how fast will she be moving at the bottom of the hill? Does her mass matter?

A \(0.25\) -hp motor is used to lift a load at the rate of \(5.0 \mathrm{~cm} / \mathrm{s}\). How great a load can it raise at this constant speed? Assume the power output of the motor to be \(0.25 \mathrm{hp}=186.5 \mathrm{~W}\). In \(1.0 \mathrm{~s}\), the load \(m g\) is lifted a distance of \(0.050 \mathrm{~m}\). Therefore, Work done in \(1.0 \mathrm{~s}=(\) weight \()(\) height change in \(1.0 \mathrm{~s})=(\mathrm{mg})(0.050 \mathrm{~m})\) By definition, Power = Work / Time, and so $$ 186.5 \mathrm{~W}=\frac{(m g)(0.050 \mathrm{~m})}{1.0 \mathrm{~s}} $$ Using \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\), we find that \(m=381 \mathrm{~kg}\). The motor can lift a load of about \(0.38 \times 10^{3} \mathrm{~kg}\) at this speed.
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