What is the work done by gravity when a 4.5 kg object is moved from start to finish point as shown in the sketch? (Use g = 10 m/s2 and round answer to two significant figures) The loop is vertical and initial motion is up 2 m 1 m 2 m finish start
What is the work done by gravity when a 4.5 kg object is moved from start to finish point as shown in the sketch? (Use g = 10 m/s2 and round answer to two significant figures) The loop is vertical and initial motion is up 2 m 1 m 2 m finish start
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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Question
![### Problem Statement
What is the work done by gravity when a 4.5 kg object is moved from start to finish point as shown in the sketch? (Use \( g = 10 \, \text{m/s}^2 \) and round answer to two significant figures)
### Diagram Description
The diagram illustrates a path taken by a 4.5 kg object in a vertical loop. The initial motion is upwards, and the object starts at a height of 2 meters. It ascends vertically another 2 meters, then moves horizontally 2 meters before descending 1 meter to reach the finish point.
### Solution Approach
1. **Determine the change in height:**
\[
\text{Initial Height} = 2 \, \text{meters}
\]
\[
\text{Final Height} = 2 + 2 - 1 = 3 \, \text{meters}
\]
2. **Calculate the change in gravitational potential energy:**
\[
\Delta h = \text{Final Height} - \text{Initial Height} = 3 - 2 = 1 \, \text{meter}
\]
3. **Use the gravitational force formula to calculate work done by gravity:**
\[
W = m \cdot g \cdot \Delta h
\]
\[
W = 4.5 \, \text{kg} \times 10 \, \text{m/s}^2 \times (0 - 1) \, \text{meter}
\]
4. **Final calculation:**
\[
W = -45 \, \text{J}
\]
5. **Rounding to two significant figures:**
\[
W = -45 \, \text{J}
\]
Thus, the work done by gravity is \(-45\) Joules.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa01d43e9-d5a5-45d6-84e7-84e66a881f35%2F8ebe522b-c393-4e87-b525-9066657ece39%2Fyqnnvgs_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
What is the work done by gravity when a 4.5 kg object is moved from start to finish point as shown in the sketch? (Use \( g = 10 \, \text{m/s}^2 \) and round answer to two significant figures)
### Diagram Description
The diagram illustrates a path taken by a 4.5 kg object in a vertical loop. The initial motion is upwards, and the object starts at a height of 2 meters. It ascends vertically another 2 meters, then moves horizontally 2 meters before descending 1 meter to reach the finish point.
### Solution Approach
1. **Determine the change in height:**
\[
\text{Initial Height} = 2 \, \text{meters}
\]
\[
\text{Final Height} = 2 + 2 - 1 = 3 \, \text{meters}
\]
2. **Calculate the change in gravitational potential energy:**
\[
\Delta h = \text{Final Height} - \text{Initial Height} = 3 - 2 = 1 \, \text{meter}
\]
3. **Use the gravitational force formula to calculate work done by gravity:**
\[
W = m \cdot g \cdot \Delta h
\]
\[
W = 4.5 \, \text{kg} \times 10 \, \text{m/s}^2 \times (0 - 1) \, \text{meter}
\]
4. **Final calculation:**
\[
W = -45 \, \text{J}
\]
5. **Rounding to two significant figures:**
\[
W = -45 \, \text{J}
\]
Thus, the work done by gravity is \(-45\) Joules.
![**Swing Dynamics and Forces**
**Scenario:**
A child is pushed in a swing as illustrated in the sketch and allowed to swing back and forth freely.
**Question:**
When the child is at the vertical position during a swing, which of the following describes the situation best?
**Diagram Explanation:**
The diagram shows a swing with a child seated, moving through three positions. The middle position is vertical, with the swing at its lowest point. The arrows indicate the velocity and the direction of the motion at different positions:
- At both ends of the swing (left and right), the child's velocity \( v \) is shown as vector arrows pointing horizontally towards the center.
- At the vertical position, the velocity also points horizontally, aligned with the direction of motion.
The forces acting on the swing include the tension in the ropes and the weight of the child, along with any centripetal force necessary to maintain the circular path.
**Options:**
- The tension force is equal to the weight plus the centripetal force.
- As the child is not moving vertically, the net vertical force is zero.
- Change in velocity vector points upward.
- The tension force is equal in magnitude to the weight.
**Correct Answer:**
The tension force is equal to the weight plus the centripetal force.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa01d43e9-d5a5-45d6-84e7-84e66a881f35%2F8ebe522b-c393-4e87-b525-9066657ece39%2Fu85yhg_processed.png&w=3840&q=75)
Transcribed Image Text:**Swing Dynamics and Forces**
**Scenario:**
A child is pushed in a swing as illustrated in the sketch and allowed to swing back and forth freely.
**Question:**
When the child is at the vertical position during a swing, which of the following describes the situation best?
**Diagram Explanation:**
The diagram shows a swing with a child seated, moving through three positions. The middle position is vertical, with the swing at its lowest point. The arrows indicate the velocity and the direction of the motion at different positions:
- At both ends of the swing (left and right), the child's velocity \( v \) is shown as vector arrows pointing horizontally towards the center.
- At the vertical position, the velocity also points horizontally, aligned with the direction of motion.
The forces acting on the swing include the tension in the ropes and the weight of the child, along with any centripetal force necessary to maintain the circular path.
**Options:**
- The tension force is equal to the weight plus the centripetal force.
- As the child is not moving vertically, the net vertical force is zero.
- Change in velocity vector points upward.
- The tension force is equal in magnitude to the weight.
**Correct Answer:**
The tension force is equal to the weight plus the centripetal force.
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