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Working Like a Horse. Your job is to lift \(30-k g\) crates a vertical distance of 0.90 \(\mathrm{m}\) from the ground onto the bed of a truck. (a) How many crates would you have to load onto the truck in 1 minute for the average power output you use to lift the crates to equal 0.50 \(\mathrm{hp} ?\) (b) How many crates for an average power output of 100 \(\mathrm{W} ?\)

Short Answer

Expert verified
For 0.50 hp, lift 85 crates; for 100 W, lift 23 crates.

Step by step solution

01

Calculate Work Done on Each Crate

First, we calculate the work done on each 30-kg crate when lifted 0.90 meters. The work done, \( W \), is given by the equation \( W = mgh \), where \( m \) is the mass in kilograms, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height in meters.\\[W = 30 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.90 \, \text{m} = 264.6 \, \text{Joules}\\]
02

Convert Power from Horsepower to Watts

Next, convert the power from horsepower to watts to simplify the calculations. 1 horsepower (hp) is equivalent to 746 watts (W). Therefore, \( 0.50 \, \text{hp} = 0.5 \times 746 \, \text{W} = 373 \, \text{W} \).
03

Calculate Number of Crates for 0.50 HP

To find the number of crates that need to be lifted in 1 minute to produce 0.50 hp, use the formula for power: \( P = \frac{W \times n}{t} \), rearranged to \( n = \frac{Pt}{W} \). Using \( P = 373 \, \text{W} \), work \( W = 264.6 \, \text{J} \), and \( t = 60 \, \text{seconds} \), calculate:\\[n = \frac{373 \, \text{W} \times 60 \, \text{s}}{264.6 \, \text{J}} \approx 84.6\] You need to lift approximately 85 crates for the average power output to equal 0.50 hp.
04

Calculate Number of Crates for 100 W

Repeat the process for the power of 100 W using the same formula \( n = \frac{Pt}{W} \). With \( P = 100 \, \text{W} \), calculate:\\[n = \frac{100 \, \text{W} \times 60 \, \text{s}}{264.6 \, \text{J}} \approx 22.7\]Hence, you would need to lift approximately 23 crates for the average power to equal 100 W.

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Key Concepts

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

Work and Energy
Work and energy are fundamental concepts in physics that help us understand how forces act and move objects. Work is defined as the energy transferred to or from an object via the application of force along a displacement. It is calculated using the formula \( W = Fd \cos\theta \), where \( W \) is the work done, \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and the displacement direction.

In this context, when lifting crates, the force applied is equal to the gravitational force, which is the mass of the object multiplied by the acceleration due to gravity. The work performed in lifting each crate is represented by the formula \( W = mgh \), where \( m \) is mass, \( g \) is gravity at approximately \( 9.8 \text{ m/s}^2 \), and \( h \) is the height lifted. Understanding these concepts allows us to calculate how much work is required to lift an object over a certain distance and how this impacts energy use.
Power Calculation
Power is the rate at which work is done or energy is transferred. It is an important measure when looking at how efficiently tasks are performed. The unit of power in the International System of Units (SI) is the watt (W), which is equivalent to one joule per second. For instance, in the problem of lifting crates, power is calculated using the formula \( P = \frac{W}{t} \), where \( P \) is power, \( W \) is work done, and \( t \) is time taken.

The exercise also involves the concept of horsepower, a unit of power commonly used historically and in everyday scenarios. One horsepower is equivalent to 746 watts. This conversion allows us to compare and measure power output effectively in different scenarios and units.

Through power calculation, we can understand the efficiency of lifting crates by determining how much power is needed over various timeframes.
Unit Conversion
Unit conversion is the process of converting a measure of physical quantities from one unit to another and is vital in solving physics problems correctly. The original problem requires converting horsepower to watts since calculations typically use the SI unit of watts for power.

Converting between units involves using conversion factors. For example, given that 1 horsepower equals 746 watts, converting 0.50 horsepower results in \( 0.50 \times 746 = 373 \text{ watts} \). This step is crucial for ensuring that all calculations are consistent, adherent to the SI units, and meaningful for further mathematical operations.

Unit conversion also helps in understanding cross-comparisons between different measurement systems, enhancing problem-solving in diverse contexts.
Mechanical Work
Mechanical work refers to the amount of energy transferred by a force acting over a distance. In mechanical systems, work is often the result of forces causing movement in physical objects. The amount of work reflects how much effort is needed, and is assessed by the formula \( W = mgh \) for an object lifted against gravity, as seen in this problem.

Understanding mechanical work helps to evaluate tasks like lifting crates, where calculating how much energy is required provides insight into how much effort a force requires to move each object over a vertical distance.
  • Work is directly proportional to both the weight of the object and the distance moved.
  • The concept of work is closely linked to energy, as it essentially reflects energy expenditure.
Grasping this concept allows students to apply it to various physical situations, deepening their understanding.
Physics Education
Physics education emphasizes the understanding of fundamental principles like work, energy, power, and unit conversion. These form the foundation of more complex physics topics and applications. For students, engaging with such concepts through practical problems enhances their comprehension and application skills.

Educational approaches often involve dissecting problems like lifting crates to teach how units and calculations are interrelated. By solving step-by-step problems, students learn critical thinking and problem-solving skills, which are valuable beyond academic settings.
  • Interactive problem-solving aids in retaining key principles.
  • Understanding real-world applications makes learning relatable.
Overall, physics education seeks to build a solid understanding that allows students to excel in diverse scientific and engineering fields.

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Most popular questions from this chapter

An elevator has mass 600 \(\mathrm{kg}\) , not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 \(\mathrm{m}\) (five floors) in 16.0 \(\mathrm{s}\) , and it is driven by a motor that can provide up to 40 \(\mathrm{hp}\) to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 \(\mathrm{kg}\).

Pushing a Cat. Your cat "Ms" (mass 7.00 \(\mathrm{kg}\) ) is trying to make it to the top of a frictionless ramp 2.00 \(\mathrm{m}\) long and inclined upward at \(30.0^{\circ}\) above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by excring a constant \(100-\mathrm{N}\) force parallel to the ramp. If Ms. takes a running start so that she is moving at 2.40 \(\mathrm{m} / \mathrm{s}\) at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.

The mass of a proton is 1836 times the mass of an electron. (a) A proton is traveling at speed \(V\) . At what speed (in terms of \(V )\) would an electron have the same kinetic energy as the proton? (b) An electron has kinetic energy \(K\) . If a proton has the same speed as the electron, what is its kinetic energy (in terms of \(K ) ?\)

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x=0\) to \(x=6.9 \mathrm{m}\) as you apply a force with \(x\) -component \(F_{x}=-[20.0 \mathrm{N}+(3.0 \mathrm{N} / \mathrm{m}) x] .\) How much work does the force you apply do on the cow during this displacement?

A Spring with Mass. We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\) , equilibrium length \(L_{0}\) , and spring constant \(k\) . The work done to stretch or compress the spring by a distance \(L\) is \(\frac{1}{2} k X^{2}\) , where \(X=L-L_{0}\) (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\) . Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v\) . (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L\) . The result is not \(\frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 \(\mathrm{kg}\) and force constant 3200 \(\mathrm{N} / \mathrm{m}\) is compressed 2.50 \(\mathrm{cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053 -kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

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