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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.

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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:
  • \( W \) 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)
This principle is key in determining how much work we need to perform to lift each crate onto the truck bed. Understanding this principle helps us calculate the energy requirements and makes solving such problems a structured process.
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.
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:
  • 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.
By following these steps, physics problems become manageable, and solutions more reachable. Structuring the problem-solving process helps in systematically reaching accurate solutions, avoiding confusion, and ensuring all aspects of the question are addressed.

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Most popular questions from this chapter

When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

A boxed 10.0-kg computer monitor is dragged by friction 5.50 m upward along a conveyor belt inclined at an angle of 36.9\(^\circ\) above the horizontal. If the monitor's speed is a constant 2.10 cm/s, how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

Two blocks are connected by a very light string passing over a massless and frictionless pulley (\(\textbf{Fig. E6.7}\)). Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string? (b) How much work is done on the 20.0-N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\), a force along the \(x\)-axis with \(x\)-component \(F_x = kx - bx^2 + cx^3\) must be applied to the free end. Here \(k = 100 \, \mathrm {N/m}\), \(b = 700 \, \mathrm {N/m{^2}}\), and \(c = 12,000 \, \mathrm{N/m}^3\). Note that \(x > 0\) when the spring is stretched and \(x< 0\) when it is compressed. How much work must be done (a) to stretch this spring by 0.050 m from its unstretched length? (b) To \(compress\) this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_x\) on \(x\). (Many real springs behave qualitatively in the same way.)

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point \(P\), it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at 12.5 m past point \(P\). (a) Use the work\(-\)energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100?

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