/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A fully loaded, slow-moving frei... [FREE SOLUTION] | 91Ó°ÊÓ

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A fully loaded, slow-moving freight elevator has a cab with a total mass of \(1200 \mathrm{~kg}\), which is required to travel upward \(54 \mathrm{~m}\) in \(3.0 \mathrm{~min},\) starting and ending at rest. The elevator's counterweight has a mass of only \(950 \mathrm{~kg}\), and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?

Short Answer

Expert verified
The average power required by the motor is 720 watts.

Step by step solution

01

Calculate the Net Force Required

First, find the net force required to lift the elevator. The gravitational force on the elevator is equal to the weight of the elevator minus the weight of the counterweight. The force of gravity on the elevator is \( F_e = m_e \cdot g \), where \( m_e = 1200 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). The force of gravity on the counterweight is \( F_c = m_c \cdot g \), where \( m_c = 950 \, \text{kg} \). The net force \( F_{net} \) can be given as:\[ F_{net} = F_e - F_c = m_e \cdot g - m_c \cdot g = (m_e - m_c) \cdot g \].Plugging in the values, \( F_{net} = (1200 - 950) \times 9.8 = 245 \times 9.8 = 2401 \, \text{N} \).
02

Calculate the Work Done by the Motor

The work done by the motor \( W \) is the product of the net force and the distance over which the force acts. Use the formula:\[ W = F_{net} \cdot d \]where \( d = 54 \, \text{m} \) is the distance over which the elevator is lifted.Substituting the known values, we get \( W = 2401 \, \text{N} \times 54 \, \text{m} = 129654 \, \text{J} \).
03

Calculate Time in Seconds

Convert the given time from minutes to seconds because power is typically expressed in seconds. The time given is \( 3.0 \, ext{min} \), which is:\[ 3.0 \, ext{min} \times 60 \, ext{s/min} = 180 \, ext{s} \].
04

Calculate Average Power Required

The average power \( P \) required by the motor is the total work done divided by the time taken. Use the power formula:\[ P = \frac{W}{t} \].Substitute \( W = 129654 \, \text{J} \) and \( t = 180 \, \text{s} \) into the formula to find the power:\[ P = \frac{129654}{180} \approx 720.3 \, \text{W} \].
05

Round the Answer

Finally, round the power to a reasonable number of significant figures, generally three digits unless otherwise specified:\[ P \approx 720 \, \text{W} \].Thus, the average power required by the motor is approximately 720 watts.

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Key Concepts

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

Understanding Net Force
Net force is the overall force acting on an object when all the individual forces are considered. In elevator mechanics, understanding the net force is crucial.
This example involves calculating the net force required to lift a freight elevator.
The elevator and counterweight create two opposing forces, influenced by gravity. Here's how it works:
  • The weight of the elevator is calculated using its mass (1200 kg) multiplied by gravity (9.8 m/s²), yielding a gravitational force of 11,760 N downward.
  • The weight of the counterweight also uses its mass (950 kg) and gravity to obtain a gravitational force of 9,310 N upward.
This scenario essentially means that the motor must overcome the difference between these two forces.
The net force (F_net) needed is given by subtracting the counterweight force from the elevator's force:\[ F_{net} = (1200 \, \text{kg} - 950 \, \text{kg}) \cdot 9.8 \, \text{m/s}^2 = 2,401 \, \text{N} \]. This net force explains how much additional effort is needed to lift the elevator upwards.
Work and Energy in Transportation
The concepts of work and energy are deeply intertwined. Work occurs when a force causes an object to move. Here, moving the elevator upwards involves work done by the motor.
The work formula is expressed as:\[ W = F_{net} \cdot d \] where \( F_{net} \) is the net force (2,401 N) and \( d \) is the distance (54 m).
  • Using the previous calculations, the work done by the motor is \( 2,401 \, \text{N} \times 54 \, \text{m} = 129,654 \, \text{J} \). This means the motor has transferred 129,654 joules of energy to move the elevator.
This energy usage underlines how much effort and energy are devoted to moving the elevator between floors. A fundamental understanding of energy considerations in mechanics helps in predicting how much force and power are necessary for specific mechanical tasks.
Deep Dive into Elevator Mechanics
Elevators are an everyday marvel of engineering, relying heavily on balanced mechanical forces. In this situation, a motor helps to counterbalance an elevator's load, providing practical insights into how elevators work.
Unlike simple lifts, real-world elevators need powerful motors and counterweights to function efficiently:
  • Counterweights reduce the load on the motor by balancing the weight of the elevator, minimizing the energy needed for lifting.
  • However, when a difference occurs in elevator and counterweight mass, the motor must supply extra force to overcome this imbalance.
Calculating average power gives us further insights into the motor's performance:The definition of average power is the rate at which work is done, given by the formula:\[ P = \frac{W}{t} \] where \( W \) is the work done (129,654 J) and \( t \) is the time taken (180 seconds).In this instance, the motor requires a power output of approximately 720 watts to lift the elevator smoothly.
These calculations demonstrate the importance of understanding how physics governs the effective design and operation of mechanical systems like elevators.

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Most popular questions from this chapter

A constant force of magnitude \(10 \mathrm{~N}\) makes an angle of \(150^{\circ}\) (measured counterclockwise) with the positive \(x\) direction as it acts on a \(2.0 \mathrm{~kg}\) object moving in an \(x y\) plane. How much work is done on the object by the force as the object moves from the origin to the point having position vector \((2.0 \mathrm{~m}) \hat{\mathrm{i}}-(4.0 \mathrm{~m}) \hat{\mathrm{j}} ?\)

61 How much work is done by a force \(\vec{F}=(2 x \mathrm{~N}) \hat{\mathrm{i}}+(3 \mathrm{~N}) \hat{\mathrm{j}},\) with \(x\) in meters, that moves a particle from a position \(\vec{r}_{i}=\) \((2 \mathrm{~m}) \hat{\mathrm{i}}+(3 \mathrm{~m}) \hat{\mathrm{j}}\) to a position \(\vec{r}_{f}=-(4 \mathrm{~m}) \hat{\mathrm{i}}-(3 \mathrm{~m}) \hat{\mathrm{j}} ?\)

A \(0.30 \mathrm{~kg}\) ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring \((k=500 \mathrm{~N} / \mathrm{m})\) whose other end is fixed. The ladle has a kinetic energy of \(10 \mathrm{~J}\) as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed \(0.10 \mathrm{~m}\) and the ladle is moving away from the equilibrium position?

What is the power of the force required to move a \(4500 \mathrm{~kg}\) elevator cab with a load of \(1800 \mathrm{~kg}\) upward at constant speed \(3.80 \mathrm{~m} / \mathrm{s} ?\)

The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

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