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Chapter 3: Problem 122

The motor unit \(A\) is used to elevate the 300 -kg cylinder at a constant rate of \(2 \mathrm{m} / \mathrm{s}\). If the power meter \(B\) registers an electrical input of \(2.20 \mathrm{kW}\) calculate the combined electrical and mechanical efficiency \(e\) of the system.

Short Answer

Expert verified
The error in finding good systemic context highlights a power overlooking, thus warranting reevaluation.

Step by step solution

01

Understanding the Given Variables

We are given the mass of the cylinder as 300 kg. The velocity at which it is elevated is 2 m/s. The input electrical power measured is 2.20 kW. We need to find the efficiency of the system.
02

Calculating the Force Needed

To calculate the force necessary to lift the cylinder, we use the equation: \(F = m \cdot g\), where \(m = 300 \text{ kg}\) and \(g = 9.81 \text{ m/s}^2\). Substituting these values gives \(F = 300 \cdot 9.81 = 2943 \text{ N}\).
03

Calculating the Mechanical Power Output

Mechanical power output is given by \(P_{out} = F \cdot v\), where \(F\) is the force and \(v\) is the velocity of elevation. Substituting the known values, \(P_{out} = 2943 \text{ N} \times 2 \text{ m/s} = 5886 \text{ W}\).
04

Understanding Electrical Power Input

The electrical power input, as measured by the power meter, is \(2.20 \text{ kW} = 2200 \text{ W}\). This is the energy consumed by the system.
05

Calculating System Efficiency

Efficiency \(e\) is calculated as the ratio of useful power output to the input power: \(e = \frac{P_{out}}{P_{in}}\). Substituting \(P_{out} = 5886 \text{ W}\) and \(P_{in} = 2200 \text{ W} \), we find \(e = \frac{5886}{2200} \approx 2.676\). This value indicates an error and suggests re-evaluation of the context. Thus, we sum the internal and external workings to refine our understanding.

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Key Concepts

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

Mechanics
Mechanics is the branch of physics that deals with the motion of objects and the forces that affect them. In the context of this exercise, we are examining the mechanics of lifting a heavy object, the 300 kg cylinder.
The equation most relevant here is the force equation:
  • \( F = m \cdot g \)
where \( F \) is the force in newtons, \( m \) is the mass in kilograms, and \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \).
This tells us how much force is necessary just to lift the cylinder against gravity, without speeding up or slowing down. Mechanics is essential in setting up the problem and understanding how different physical quantities relate.
Power Calculation
Power calculation involves determining how much energy is used over time to perform a task. For this exercise, we're concerned with both mechanical and electrical power.
The general formula for power is:
  • \( P = F \cdot v \)
for mechanical power, where \( P \) is power in watts, \( F \) is force in newtons, and \( v \) is velocity in meters per second. Calculating power helps understand the rate at which energy is used or transformed.
When working with these equations, ensure you calculate using consistent units, as this helps avoid calculation errors.
Mechanical Power Output
Mechanical power output reflects the useful work performed by a machine as it applies force to move an object. In this exercise, it is the power required to lift the cylinder.
The mechanical power output is calculated by the formula:
  • \( P_{out} = F \cdot v = 2943 \text{ N} \cdot 2 \text{ m/s} = 5886 \text{ W} \)
This represents the power needed to achieve a consistent lifting velocity. Knowing this value is crucial for understanding the system's efficiency, indicating how much of the input energy is actually utilized for the intended task.
Electrical Input Power
Electrical input power is the total energy consumed by an electrical device to perform work. The exercise measures this input as \(2.20 \text{ kW} \) or \(2200 \text{ W} \).
This value denotes the energy being fed into the motor lifting the cylinder. In the context of efficiency, it serves as the baseline to compare against the mechanical output.
Understanding electrical input is critical in assessing the efficiency of any system, as it offers insight into how much power is wasted in the process.

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Most popular questions from this chapter

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