Chapter 6: Problem 57
Your job is to lift 30-kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. How many crates would you have to load onto the truck in 1 minute (a) for the average power output you use to lift the crates to equal 0.50 hp; (b) for an average power output of 100 W?
Short Answer
Expert verified
For 0.50 hp, load 84 crates; for 100 W, load 22 crates.
Step by step solution
01
Understand the Problem
We need to calculate the number of 30-kg crates lifted onto a truck in 1 minute with specified power outputs. Power is defined as work done per unit time. We will consider two different power outputs: 0.50 horsepower (hp) and 100 watts (W).
02
Convert Horsepower to Watts
Since power output is given in horsepower, we first convert 0.50 hp to watts. Using the conversion factor, 1 hp = 745.7 watts. Thus, 0.50 hp = 0.50 * 745.7 W = 372.85 W.
03
Calculate Work Done Per Crate
The work done to lift one crate is calculated using the formula: \( W = mgh \), where \( m = 30 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \) (acceleration due to gravity), and \( h = 0.90 \text{ m} \). Thus \( W = 30 \times 9.8 \times 0.90 = 264.6 \text{ J} \).
04
Calculate Number of Crates for 0.50 hp Output
To find the number of crates for 0.50 hp, divide the power in watts by the work per crate and multiply by time in seconds (since \( P = \frac{W}{t} \)). \( \frac{372.85 \text{ W} \times 60 \text{ s}}{264.6 \text{ J}} \approx 84.48 \). Since you can't lift a fraction of a crate, you would need to load 84 crates.
05
Calculate Number of Crates for 100 W Output
For 100 W output, use the same method: \( \frac{100 \text{ W} \times 60 \text{ s}}{264.6 \text{ J}} \approx 22.68 \) crates. Rounding down, you can load 22 crates within a minute since partial crates can't be lifted.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Work-energy principle
The work-energy principle is a fundamental concept in physics that connects the work done on an object to the change in its kinetic energy. Simply put, when work is done on an object by applying a force, the object gains energy. This energy is often in the form of kinetic energy, but depending on the scenario, it can also be potential or mechanical energy.
In our exercise, we are lifting crates, which involves converting potential energy due to gravity into mechanical work. The work done to lift one crate is calculated using the formula \( W = mgh \), where:
In our exercise, we are lifting crates, which involves converting potential energy due to gravity into mechanical work. The work done to lift one crate is calculated using the formula \( W = mgh \), where:
- \( W \) is the work done (in joules)
- \( m \) is the mass of the crate (in kilograms)
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²)
- \( h \) is the height lifted (in meters)
Power conversion
Power conversion is about translating different units of power to understand and compare their values across different measuring systems. Power, at its core, is the rate at which work is done or energy is transferred over time. The common units for measuring power include watts (W) and horsepower (hp).
In our problem, the power requirements are given both in horsepower and watts. We need to convert horsepower to watts for consistency. This conversion uses the factor that 1 horsepower equals 745.7 watts. For example, the problem states an output of 0.50 hp, which translates to 372.85 W. Using consistent units allows us to easily calculate the number of crates that can be lifted within the power constraints.
This conversion process is crucial for physics problems since it ensures that all computations are accurate and valid, especially when involving power-related calculations.
In our problem, the power requirements are given both in horsepower and watts. We need to convert horsepower to watts for consistency. This conversion uses the factor that 1 horsepower equals 745.7 watts. For example, the problem states an output of 0.50 hp, which translates to 372.85 W. Using consistent units allows us to easily calculate the number of crates that can be lifted within the power constraints.
This conversion process is crucial for physics problems since it ensures that all computations are accurate and valid, especially when involving power-related calculations.
Physics problem solving
Approaching physics problems requires a structured problem-solving method that incorporates understanding the problem, selecting the appropriate equations, and applying them correctly. A strong grasp of underlying principles like work, energy, and power is essential.
Our example problem involves a straightforward application of physics principles to determine how many crates can be lifted at specified power outputs. Breaking it down step-by-step:
Our example problem involves a straightforward application of physics principles to determine how many crates can be lifted at specified power outputs. Breaking it down step-by-step:
- Identify what needs to be calculated (number of crates).
- Use relevant formulas: work done (\( W = mgh \)) and power formula \( P = \frac{W}{t} \).
- Convert any necessary units for consistency (like converting horsepower to watts).
- Plug in known values and solve step by step for each power requirement.