Initial Height h₁ m = 325 kg h₁ = 19.5 m 1 Car mass, with passengers = m No friction. 2 Bottom 3 Final Height h₂ h2=4.7 m A roller coaster car of mass m is rolling down the track, starting from rest, as shown in the figure. The initial height from the ground is h1 (Position-1), while the final height is h2 (Position-3). The system has no friction. What is the speed of the car when it reaches the bottom (Position-2) of the track.

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)...
icon
Related questions
icon
Concept explainers
Question
### Energy Conservation in a Roller Coaster Ride

#### Diagram Explanation:

A diagram of a roller coaster track is shown, divided into three key positions:

1. **Position 1 (Initial Height, \( h_1 \))**:
   - The roller coaster starts at the topmost point of the track with an initial height \( h_1 = 19.5 \) meters from the ground.
   
2. **Position 2 (Bottom)**:
   - The lowest point of the track, where we will calculate the speed of the roller coaster.
   
3. **Position 3 (Final Height, \( h_2 \))**:
   - Another elevated point on the track where the final height \( h_2 \) is given as 4.7 meters.
   
- **Roller Coaster Car Information**:
   - The mass of the roller coaster car, including passengers, is denoted as \( m = 325 \) kg.
   - The track has no friction, implying that mechanical energy is conserved throughout the ride.

#### Problem Statement:

A roller coaster car of mass \( m \) is rolling down the track, starting from rest, as shown in the figure. The initial height from the ground is \( h_1 \) (Position-1), while the final height is \( h_2 \) (Position-3). The system has no friction.

What is the speed of the car when it reaches the bottom (Position-2) of the track?

#### Given:
- \( m = 325 \) kg
- \( h_1 = 19.5 \) m
- \( h_2 = 4.7 \) m

#### Solution Approach:

Using the conservation of mechanical energy principle:

1. **Potential Energy at Initial Point (Position 1)**:
   - \( PE_1 = m \cdot g \cdot h_1 \)

2. **Kinetic Energy at Initial Point (Position 1)**:
   - Since the car starts from rest, \( KE_1 = 0 \)

3. **Potential Energy at Bottom (Position 2)**:
   - At the bottom, \( h = 0 \), so \( PE_2 = 0 \)

4. **Kinetic Energy at Bottom (Position 2)**:
   - \( KE_2 = \frac{1}{2} m v^2 \), where \( v \)
Transcribed Image Text:### Energy Conservation in a Roller Coaster Ride #### Diagram Explanation: A diagram of a roller coaster track is shown, divided into three key positions: 1. **Position 1 (Initial Height, \( h_1 \))**: - The roller coaster starts at the topmost point of the track with an initial height \( h_1 = 19.5 \) meters from the ground. 2. **Position 2 (Bottom)**: - The lowest point of the track, where we will calculate the speed of the roller coaster. 3. **Position 3 (Final Height, \( h_2 \))**: - Another elevated point on the track where the final height \( h_2 \) is given as 4.7 meters. - **Roller Coaster Car Information**: - The mass of the roller coaster car, including passengers, is denoted as \( m = 325 \) kg. - The track has no friction, implying that mechanical energy is conserved throughout the ride. #### Problem Statement: A roller coaster car of mass \( m \) is rolling down the track, starting from rest, as shown in the figure. The initial height from the ground is \( h_1 \) (Position-1), while the final height is \( h_2 \) (Position-3). The system has no friction. What is the speed of the car when it reaches the bottom (Position-2) of the track? #### Given: - \( m = 325 \) kg - \( h_1 = 19.5 \) m - \( h_2 = 4.7 \) m #### Solution Approach: Using the conservation of mechanical energy principle: 1. **Potential Energy at Initial Point (Position 1)**: - \( PE_1 = m \cdot g \cdot h_1 \) 2. **Kinetic Energy at Initial Point (Position 1)**: - Since the car starts from rest, \( KE_1 = 0 \) 3. **Potential Energy at Bottom (Position 2)**: - At the bottom, \( h = 0 \), so \( PE_2 = 0 \) 4. **Kinetic Energy at Bottom (Position 2)**: - \( KE_2 = \frac{1}{2} m v^2 \), where \( v \)
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Moment of inertia
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.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
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…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON