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Chapter 5: Problem 58

A 650-kg elevator starts from rest and moves upward for \(3.00 \mathrm{~s}\) with constant acceleration until it reaches its cruising speed, \(1.75 \mathrm{~m} / \mathrm{s}\). (a) What is the average power of the elevator motor during this period? (b) How does this amount of power compare with its power during an upward trip with constant speed?

Short Answer

Expert verified
The average power of the elevator motor during acceleration is higher than the power consumption at constant speed. This is expected, as additional power is required to overcome inertia and move the elevator from rest until it reaches its cruising speed.

Step by step solution

01

Calculate the Final Velocity

Use the standard formulation: \(v_{f} = v_{i} + at\). Here, \(a\) is the constant acceleration and \(t\) is the time. Given that the final velocity \(v_{f}\) is 1.75 m/s, initial velocity \(v_{i}\) is 0 (as the elevator starts from rest), and time \(t\) is 3 seconds. You can solve it for acceleration: \(a = (v_{f} - v_{i}) / t\)
02

Calculate the Work Done

Work done can be calculated using the equation: \(W = F \cdot d\). Here, \(F\) is the force and \(d\) is the displacement. Force \(F\) is the vector sum of the elevator's weight and the net force: \(F = m \cdot g + m \cdot a\). Here, \(m\) is the mass of the elevator, \(g\) is the acceleration due to gravity (9.8 m/s\(^2\)), and \(a\) is the acceleration calculated in step 1. Displacement \(d\) equals the product of \(v_{f}\) and \(t/2\). Now, plug these values in to calculate the work.
03

Calculate the Average Power

Power can be determined by the equation: \(P = W / t\). Here, \(W\) is the work and \(t\) is the time. Plug in the values calculated in step 2 to find the average power during the acceleration period.
04

Calculate the Power during Constant Motion

During constant motion, there's no acceleration. Forces acting on the elevator are balanced. Hence, the force in operation is the weight of the elevator: \(F = m \cdot g\). Work done \(W\) is then: \(F \cdot d\), where displacement \(d\) is the same distance as in the previous example, i.e., \(v_{f} \cdot t / 2\). The power can be found from: \(P = W / t\).
05

Compare the Powers

Now compare the power found in steps 3 and 4 to complete part (b) of the problem. This will indicate how the power consumption changes when the elevator is accelerating compared to when it travels at a constant speed.

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Key Concepts

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

Average Power
Average power is a concept that describes the rate at which work is done over a period of time. In physics, power is the amount of work performed or energy converted per unit of time. For an elevator moving upwards, average power helps us understand how much energy is required to lift the elevator over a certain distance under the influence of gravity.

The formula for average power (\(P\)) is given by:
  • \(P = \frac{W}{t},\) where \(W\) is the work done and \(t\) is the time taken.
For an elevator, the work done is the force applied to move up against gravity, multiplied by the distance traveled. By calculating average power, we gain insights into the energy efficiency and performance of the elevator motor during its operation.
Elevator Motion
Elevator motion primarily involves the movement of an elevator up or down a shaft, usually with the assistance of an electric motor. Handling such movements includes understanding concepts like acceleration, constant speed, and the work-energy principle.

When an elevator starts moving from rest, as in our problem, it undergoes an acceleration to reach its cruising speed. During this phase, the elevator's motion is governed by the principles of physics that determine how much force and power are necessary to achieve the desired speed.
  • Motion starts from rest and transitions to a constant speed.
  • Involves applying force to overcome gravitational pull.
  • Considers the energy required to lift the elevator against gravity and friction.
Understanding elevator motion is critical in calculating the power necessary for different phases of its movement, such as acceleration and constant speed.
Constant Acceleration
Constant acceleration is a condition in which the acceleration of an object remains unchanged over a period of time. This situation simplifies calculations related to motion, such as velocity and displacement.

In this context, the elevator starts from rest and moves upward with constant acceleration until reaching a specified speed. The formulas used here can help understand the elevator's journey.
  • Starting from rest means initial velocity (\(v_i\)) is 0.
  • Final velocity (\(v_f\)) can be calculated using \(v_f = v_i + a \times t\).
  • Once acceleration is determined, it is easier to compute the work done and power used.
Knowing how constant acceleration affects motion aids in predicting and controlling the performance of moving objects like elevators.
Work and Energy
Work and energy are fundamental concepts in physics that describe the capacity to do work and the actual accomplishment of work, respectively. In the context of an elevator, these principles apply to understanding how energy is used to lift the elevator.

Work (\(W\)) in physics is calculated as the product of the force applied to an object and the displacement caused by that force. For an elevator:
  • Work done is the force moving the elevator upwards (\(W = F \cdot d\)
  • Force \(F\) includes both the weight (\(m \cdot g\)) and any additional force needed for acceleration.
  • Energy required is directly related to the work done by the motor.
These concepts underpin the determination of power since energy expended (or work done) over time gives the power鈥攁 crucial aspect for evaluating the performance of the elevator's motor.

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Most popular questions from this chapter

The force acting on an object is given by \(F_{x}=\) \((8 x-16) \mathrm{N}\), where \(x\) is in meters. (a) Make a plot of this force versus \(x\) from \(x=0\) to \(x=3.00 \mathrm{~m}\). (b) From your graph, find the net work done by the force as the object moves from \(x=0\) to \(x=3.00 \mathrm{~m}\).

\(\mathrm{S}\) A projectile of mass \(m\) is fired horizontally with an initial speed of \(v_{0}\) from a height of \(h\) above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of \(m, v_{0}, h\), and \(g\) : (a) the work done by the force of gravity on the projectile, (b) the change in kinetic energy of the projectile since it was fired, and (c) the final kinetic energy of the projectile. (d) Are any of the answers changed if the initial angle is changed?

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of \(4.0 \mathrm{~m} / \mathrm{s}\) up a \(20^{\circ}\) inclined track. The combined mass of monkey and sled is \(20 \mathrm{~kg}\), and the coefficient of kinetic friction between sled and incline is \(0.20\). How far up the incline do the monkey and sled move?

A daredevil wishes to bungee-jump from a hot-air balloon \(65.0 \mathrm{~m}\) above a carnival midway. He will use a piece of uniform elastic cord tied to a harness around his body to stop his fall at a point \(10.0 \mathrm{~m}\) above the ground. Model his body as a particle and the cord as having negligible mass and a tension force described by Hooke's force law. In a preliminary test, hanging at rest from a \(5.00-\mathrm{m}\) length of the cord, the jumper finds that his body weight stretches it by \(1.50 \mathrm{~m}\). He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. (a) What length of cord should he use? (b) What maximum acceleration will he experience?

A weight lifter lifts a \(350-N\) set of weights from ground level to a position over his head, a vertical distance of \(2.00 \mathrm{~m}\). How much work does the weight lifter do, assuming he moves the weights at constant speed?
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