A 25.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 43.0 from the vertical and releases her from rest.
A 25.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 43.0 from the vertical and releases her from rest.
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)...
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7.10
![### Potential Energy and Kinetic Energy in a Swing System
**Scenario:**
A child with a mass of 25.0 kg plays on a swing with support ropes that are 2.10 meters long. A friend pulls the child back until the ropes form an angle of 43.0° from the vertical and then releases her from rest.
---
**Part A: Potential Energy Calculation**
**Question:**
What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?
**Equation Input:**
\[ U = \ \_\_\_\_\_\ \text{J} \]
\[ \boxed{Submit} \]
\( \boxed{Request\ Answer} \)
Explanation: Calculate the potential energy at the initial position (where she is pulled back) using the formula
\[ U = mgh \]
where \( m \) is the mass (25.0 kg), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the vertical height from the lowest point of the swing.
---
**Part B: Speed at the Bottom of the Swing**
**Question:**
How fast will she be moving at the bottom of the swing?
**Equation Input:**
\[ v = \ \_\_\_\_\_\ \text{m/s} \]
\[ \boxed{Submit} \]
\[ \boxed{Request\ Answer} \]
Explanation: Use the principle of conservation of energy. At the bottom of the swing, all the potential energy will be converted into kinetic energy. The speed can be found using:
\[ \frac{1}{2}mv^2 = mgh \]
Solving for \( v \) gives:
\[ v = \sqrt{2gh} \]
---
**Part C: Work Done by Tension**
**Question:**
How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
**Equation Input:**
\[ W = \ \_\_\_\_\_\ \text{J} \]
\[ \boxed{Submit} \]
\[ \boxed{Request\ Answer} \]
Explanation: The work done by the tension in the ropes as the child swings from the initial position to the bottom is zero because tension acts perpendicular to the direction of the child's motion and does not do any work.
---
This exercise illustrates the concepts of potential energy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddf14729-dbf2-4574-b793-bebd8df9c378%2F9b6b6afc-0424-4a05-8f4c-03af1fd60efb%2Fcb09imc_processed.png&w=3840&q=75)
Transcribed Image Text:### Potential Energy and Kinetic Energy in a Swing System
**Scenario:**
A child with a mass of 25.0 kg plays on a swing with support ropes that are 2.10 meters long. A friend pulls the child back until the ropes form an angle of 43.0° from the vertical and then releases her from rest.
---
**Part A: Potential Energy Calculation**
**Question:**
What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?
**Equation Input:**
\[ U = \ \_\_\_\_\_\ \text{J} \]
\[ \boxed{Submit} \]
\( \boxed{Request\ Answer} \)
Explanation: Calculate the potential energy at the initial position (where she is pulled back) using the formula
\[ U = mgh \]
where \( m \) is the mass (25.0 kg), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the vertical height from the lowest point of the swing.
---
**Part B: Speed at the Bottom of the Swing**
**Question:**
How fast will she be moving at the bottom of the swing?
**Equation Input:**
\[ v = \ \_\_\_\_\_\ \text{m/s} \]
\[ \boxed{Submit} \]
\[ \boxed{Request\ Answer} \]
Explanation: Use the principle of conservation of energy. At the bottom of the swing, all the potential energy will be converted into kinetic energy. The speed can be found using:
\[ \frac{1}{2}mv^2 = mgh \]
Solving for \( v \) gives:
\[ v = \sqrt{2gh} \]
---
**Part C: Work Done by Tension**
**Question:**
How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
**Equation Input:**
\[ W = \ \_\_\_\_\_\ \text{J} \]
\[ \boxed{Submit} \]
\[ \boxed{Request\ Answer} \]
Explanation: The work done by the tension in the ropes as the child swings from the initial position to the bottom is zero because tension acts perpendicular to the direction of the child's motion and does not do any work.
---
This exercise illustrates the concepts of potential energy
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