Calculate the energy required to lift an elevator and its occupants with amass of 2000 kg for the following situations: (a) between the first and thesecond floor, (b) between the third and the fourth floor, (c) between the firstand the fourth floor. (as shown .) The vertical distance between eachfloor is 4.5 m. We can use as shown to analyze this problem; the energy required to lift the elevator is equal to the change in its potential energy, starting with 4Ah = 4.5 mAh = 4.5 m2Ah = 4.5 mFIGURE 1 3.3 The work required to lift an object with a mass m by a vertical distance Ahis called gravitational potential energy. It is the mechanical work that mustbe performed to overcome the gravitational pull of the earth on the object(see Figure 13.2). The change in the potential energy of the object when itselevation is changed is given bychange in potential energy = APE = mg Ah13.4We can use Equation (13.4) to analyze this problem; the energy requiredto lift the elevator is equal to the change in its potential energy, starting with(a) change in potential energy = mg Ah= (200 kg) 9.81(4.5 m) = 88,290 J(b) change in potential energy = mg Ah= (200 kg)|9.81 (4.5 m) = 88,290 J(c) change in potential energy = mg Ah(200 kg) 9.81(13.5 m) = 264, 870 JNote that the amount of energy required to lift the elevator from the firstto the second floor and from the third to the fourth floor is the same. Alsorealize that we have neglected any frictional effect in our analysis. The actualenergy requirement would be greater in the presence of frictional effect.

Given data: mass of elevator = 2000 kg height of each story = 4.5 m

Answered: 4 Ah = 4.5 m Ah = 4.5 m 2 Ah = 4.5 m…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Question

Calculate the energy required to lift an elevator and its occupants with a
mass of 2000 kg for the following situations: (a) between the first and the
second floor, (b) between the third and the fourth floor, (c) between the first
and the fourth floor. (as shown .) The vertical distance between each
floor is 4.5 m. We can use as shown to analyze this problem; the energy required to lift the elevator is equal to the change in its potential energy, starting with

4
Ah = 4.5 m
Ah = 4.5 m
2
Ah = 4.5 m
FIGURE 1 3.3

The work required to lift an object with a mass m by a vertical distance Ah
is called gravitational potential energy. It is the mechanical work that must
be performed to overcome the gravitational pull of the earth on the object
(see Figure 13.2). The change in the potential energy of the object when its
elevation is changed is given by
change in potential energy = APE = mg Ah
13.4
We can use Equation (13.4) to analyze this problem; the energy required
to lift the elevator is equal to the change in its potential energy, starting with
(a) change in potential energy = mg Ah
= (200 kg) 9.81(4.5 m) = 88,290 J
(b) change in potential energy = mg Ah
= (200 kg)|9.81 (4.5 m) = 88,290 J
(c) change in potential energy = mg Ah
(200 kg) 9.81(13.5 m) = 264, 870 J
Note that the amount of energy required to lift the elevator from the first
to the second floor and from the third to the fourth floor is the same. Also
realize that we have neglected any frictional effect in our analysis. The actual
energy requirement would be greater in the presence of frictional effect.

Expert Solution

Step by step

Solved in 4 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions