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Chapter 13: Problem 47

It takes 15 s to raise a \(1200-\mathrm{kg}\) car and the supporting \(300-\mathrm{kg}\) hydraulic car-lift platform to a height of \(2.8 \mathrm{m}\). Determine \((a)\) the average output power delivered by the hydraulic pump to lift the system, \((b)\) the average electric power required, knowing that the overall conversion efficiency from electric to mechanical power for the system is 82 percent.

Short Answer

Expert verified
(a) 2744 W, (b) 3346 W.

Step by step solution

01

Calculate the Gravitational Force

The gravitational force (weight) of an object can be calculated using Earth鈥檚 gravitational pull, given by the formula: \( F = m \cdot g \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately \( 9.8 \ \text{m/s}^2 \)). Here, the total mass \( m \) is the sum of the car and the platform, i.e., \( 1200 + 300 = 1500 \ \text{kg} \). Thus, the gravitational force \( F \) is: \[ F = 1500 \times 9.8 = 14700 \ \text{N} \]
02

Compute the Work Done

Work is the product of force and distance: \( W = F \cdot d \). Here, \( F = 14700 \ \text{N} \) and the distance \( d = 2.8 \ \text{m} \). Therefore, the work done \( W \) is: \[ W = 14700 \times 2.8 = 41160 \ \text{J} \]
03

Calculate Average Output Power

Power is defined as work done over time, \( P = \frac{W}{t} \). Given that the work \( W = 41160 \ \text{J} \) and the time \( t = 15 \ \text{s} \), the average output power \( P \) is: \[ P = \frac{41160}{15} = 2744 \ \text{W} \]
04

Determine Average Electric Power Required

The average electric power required can be found by considering the efficiency of the system. Given that the efficiency \( \eta = 82\% = 0.82 \) and the output power \( P = 2744 \ \text{W} \), the input electric power \( P_{\text{in}} \) is: \[ P_{\text{in}} = \frac{P}{\eta} = \frac{2744}{0.82} \approx 3346 \ \text{W} \]

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Key Concepts

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

Gravitational Force
Gravitational force is the force with which the Earth pulls objects towards itself. It is essential in problems involving lifting or supporting objects. We calculate this force using the formula: \( F = m \cdot g \). Here, \( m \) is the total mass and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \). For our problem, the total mass of the car and the platform is \( 1500 \text{ kg} \).
This results in a gravitational force:
  • \( F = 1500 \times 9.8 = 14700 \text{ N} \)
Gravitational force is crucial for calculating the work done when lifting an object. Knowing how to compute this force gives us a deeper understanding of the forces at play in mechanical systems.
Work Done
Work done is a concept that represents the energy transferred to or from an object via a force acting over a distance. To calculate the work done when lifting an object, we use the formula \( W = F \cdot d \), where \( F \) is the force applied and \( d \) is the distance moved. In our example, the gravitational force is \( 14700 \text{ N} \) and the height lifted is \( 2.8 \text{ m} \).
  • \( W = 14700 \times 2.8 = 41160 \text{ J} \)
Work is measured in joules (J) and represents how much energy has been used. Thorough understanding of work done helps in predicting how much effort and energy is required for mechanical tasks.
Electric Power
Electric power is the rate at which energy is used or transferred by an electric circuit. It is crucial to understand when converting mechanical tasks into electrical terms. Power is measured in watts (W) and calculated using the formula \( P = \frac{W}{t} \), with \( W \) being the work done and \( t \) the time.
In our problem, the work done was \( 41160 \text{ J} \) over \( 15 \text{ s} \). This gives an average output power:
  • \( P = \frac{41160}{15} = 2744 \text{ W} \)
Understanding electric power is vital for determining how much energy input is required for a task and how efficient an electric system is.
Efficiency Calculation
Efficiency is a measure of how well energy is converted from one form to another without losses. It is calculated by dividing the useful output energy by the input energy. In this exercise, the mechanical system efficiency is given as 82%, expressed as a decimal \( \eta = 0.82 \).
The formula to find the required input electric power \( P_{\text{in}} \) is:
  • \( P_{\text{in}} = \frac{P}{\eta} \)
Given the output power \( P = 2744 \text{ W} \), the input electric power required is:
  • \( P_{\text{in}} = \frac{2744}{0.82} \approx 3346 \text{ W} \)
Understanding efficiency allows us to optimize systems for better energy use, resulting in cost and energy savings.

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Most popular questions from this chapter

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