An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession. The elevator accelerates upward from rest at a rate of 1 ms2 for 1.9 s. (ii) Calculate the tension in the cable supporting the elevator. (iv) Calculate the velocity of the elevator after this time. The elevator continues upward at constant velocity for 8.8 s. (iii) How high has the elevator moved during this time?
An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession. The elevator accelerates upward from rest at a rate of 1 ms2 for 1.9 s. (ii) Calculate the tension in the cable supporting the elevator. (iv) Calculate the velocity of the elevator after this time. The elevator continues upward at constant velocity for 8.8 s. (iii) How high has the elevator moved during this time?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Question
An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession.
The elevator accelerates upward from rest at a rate of 1 ms2 for 1.9 s.
(ii) Calculate the tension in the cable supporting the elevator.
(iv) Calculate the velocity of the elevator after this time.
The elevator continues upward at constant velocity for 8.8 s.
(iii) How high has the elevator moved during this time?
(c) The elevator decelerates at a rate of 0.7 ms2 for 2.85 s.
(ii) Calculate the tension in the cable supporting the elevator.
(iii) How high has the elevator moved during this time?
(d) What is the total distance the elevator has moved up?
![**Problem 1: Elevator Motion Analysis**
An elevator filled with passengers has a total mass of 1850 kg. The elevator undergoes the following motions sequentially from (a) to (c).
**Diagram Explanation:**
- The diagram shows an elevator with a tension force (T) acting upward and a gravitational force (mg) acting downward.
- A person is shown inside the elevator.
**Instructions:**
For each part below, draw a free body diagram of the elevator in your notebook and indicate the acceleration and velocity vectors.
**Coordinate System:**
- \( y \) axis: vertical direction
- \( x \) axis: horizontal direction
**Parts (a) to (c):**
**(a) Description:**
- The elevator accelerates upward from rest at 1 m/s² for 1.9 seconds.
**Equations and Calculations:**
(i) **Newton’s Second Law in the y-direction:**
\[
\Sigma F_y = T - m_E g = m_E a
\]
- Here, \( m_E = 1850 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( a = 1 \text{ m/s}^2 \)
(ii) **Tension in the cable:**
\[
T = 18130 \text{ N}
\]
(iii) **Displacement of the elevator:**
\[
\Delta y = 1.8 \text{ m}
\]
(iv) **Velocity of the elevator at end of acceleration:**
\[
v(t=1.9 \text{ s}) = 0.95 \text{ m/s}
\]
**(b) Description:**
- The elevator continues upward at constant velocity for 8.8 seconds.
**Newton’s Law in the y-direction:**
- \( T = m_E g \), indicating balanced forces as the acceleration is zero during this phase.
_Note: The solution involves using physics principles like Newton’s law to calculate forces and kinematic equations to determine displacement and velocity._](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe12e5cd-e451-4d34-8110-b6d444fb2efb%2F7832ce86-e22d-4d01-bd3f-b917b32d16dd%2F7r76a1b_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 1: Elevator Motion Analysis**
An elevator filled with passengers has a total mass of 1850 kg. The elevator undergoes the following motions sequentially from (a) to (c).
**Diagram Explanation:**
- The diagram shows an elevator with a tension force (T) acting upward and a gravitational force (mg) acting downward.
- A person is shown inside the elevator.
**Instructions:**
For each part below, draw a free body diagram of the elevator in your notebook and indicate the acceleration and velocity vectors.
**Coordinate System:**
- \( y \) axis: vertical direction
- \( x \) axis: horizontal direction
**Parts (a) to (c):**
**(a) Description:**
- The elevator accelerates upward from rest at 1 m/s² for 1.9 seconds.
**Equations and Calculations:**
(i) **Newton’s Second Law in the y-direction:**
\[
\Sigma F_y = T - m_E g = m_E a
\]
- Here, \( m_E = 1850 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), \( a = 1 \text{ m/s}^2 \)
(ii) **Tension in the cable:**
\[
T = 18130 \text{ N}
\]
(iii) **Displacement of the elevator:**
\[
\Delta y = 1.8 \text{ m}
\]
(iv) **Velocity of the elevator at end of acceleration:**
\[
v(t=1.9 \text{ s}) = 0.95 \text{ m/s}
\]
**(b) Description:**
- The elevator continues upward at constant velocity for 8.8 seconds.
**Newton’s Law in the y-direction:**
- \( T = m_E g \), indicating balanced forces as the acceleration is zero during this phase.
_Note: The solution involves using physics principles like Newton’s law to calculate forces and kinematic equations to determine displacement and velocity._
![**Section (a): Accelerating Upward**
The elevator accelerates upward from rest at a rate of \(1 \, \text{m/s}^2\) for 1.9 seconds.
1. **Newton’s Second Law in the y-direction**:
\[
\sum F_y = -1 \rightarrow m_e g + 1 \rightarrow T = 1 \rightarrow m_e a
\]
2. **Calculate the tension in the cable supporting the elevator**:
\[
T = 18130 \, \text{N}
\]
3. **How high has the elevator moved during this time?**
\[
\Delta y = 1.8 \, \text{m}
\]
4. **Calculate the velocity of the elevator after this time**:
\[
v(t = 1.9 \, \text{s}) = 0.95 \, \text{m/s}
\]
**Section (b): Constant Velocity**
The elevator continues upward at a constant velocity for 8.8 seconds.
1. **Newton's Law in the y-direction**:
\[
\sum F_y = -1 \rightarrow m_e g + 1 \rightarrow T = -1 \rightarrow m_e a
\]
2. **Calculate the tension in the cable supporting the elevator**:
\[
T = 18130 \, \text{N}
\]
3. **How high has the elevator moved during this time?**
\[
\Delta y = 38.72 \, \text{m}
\]
**Section (c): Decelerating**
The elevator decelerates at a rate of \(0.7 \, \text{m/s}^2\) for 2.85 seconds.
1. **Newton’s Law in the y-direction**:
\[
\sum F_y = -1 \rightarrow m_e g + 1 \rightarrow T = -1 \rightarrow m_e a
\]
2. **Calculate the tension in the cable supporting the elevator**:
\[
T = 18130 \, \text{N}
\]
3. **How high has the elevator moved during this time?**
\[
\Delta y = 1.99 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe12e5cd-e451-4d34-8110-b6d444fb2efb%2F7832ce86-e22d-4d01-bd3f-b917b32d16dd%2Fcsu9s8s_processed.png&w=3840&q=75)
Transcribed Image Text:**Section (a): Accelerating Upward**
The elevator accelerates upward from rest at a rate of \(1 \, \text{m/s}^2\) for 1.9 seconds.
1. **Newton’s Second Law in the y-direction**:
\[
\sum F_y = -1 \rightarrow m_e g + 1 \rightarrow T = 1 \rightarrow m_e a
\]
2. **Calculate the tension in the cable supporting the elevator**:
\[
T = 18130 \, \text{N}
\]
3. **How high has the elevator moved during this time?**
\[
\Delta y = 1.8 \, \text{m}
\]
4. **Calculate the velocity of the elevator after this time**:
\[
v(t = 1.9 \, \text{s}) = 0.95 \, \text{m/s}
\]
**Section (b): Constant Velocity**
The elevator continues upward at a constant velocity for 8.8 seconds.
1. **Newton's Law in the y-direction**:
\[
\sum F_y = -1 \rightarrow m_e g + 1 \rightarrow T = -1 \rightarrow m_e a
\]
2. **Calculate the tension in the cable supporting the elevator**:
\[
T = 18130 \, \text{N}
\]
3. **How high has the elevator moved during this time?**
\[
\Delta y = 38.72 \, \text{m}
\]
**Section (c): Decelerating**
The elevator decelerates at a rate of \(0.7 \, \text{m/s}^2\) for 2.85 seconds.
1. **Newton’s Law in the y-direction**:
\[
\sum F_y = -1 \rightarrow m_e g + 1 \rightarrow T = -1 \rightarrow m_e a
\]
2. **Calculate the tension in the cable supporting the elevator**:
\[
T = 18130 \, \text{N}
\]
3. **How high has the elevator moved during this time?**
\[
\Delta y = 1.99 \
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON