College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Problem 9:**
To cheer him up, someone hands Mikey (still 27.0 kg) a helium-filled balloon. The scale now reads 25.0 kg. What is the force exerted by the balloon? (Include a free-body diagram.)
**Explanation:**
To solve this problem, we need to calculate the force exerted by the helium-filled balloon, which causes the scale reading to drop from Mikey's weight of 27.0 kg to 25.0 kg.
**Steps:**
1. Determine the difference in scale reading:
\[
\text{Difference in weight} = 27.0 \, \text{kg} - 25.0 \, \text{kg} = 2.0 \, \text{kg}
\]
2. Calculate the force exerted by the balloon (which is the buoyant force reducing Mikey’s apparent weight):
\[
F_{\text{balloon}} = \text{Difference in weight} \times g
\]
where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
3. \[
F_{\text{balloon}} = 2.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N}
\]
**Free-Body Diagram:**
The free-body diagram would show:
- Downward force:
- Mikey’s weight: \( 27.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 264.6 \, \text{N} \)
- Upward force:
- Buoyant force by balloon: \( 19.6 \, \text{N} \)
In the diagram, these forces balance out to reflect the apparent weight of 25.0 kg on the scale:
- Apparent weight: \( 25.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 245.0 \, \text{N} \)
This illustrates the decrease in the weight reading due to the upward buoyant force exerted by the helium balloon.](https://content.bartleby.com/qna-images/question/4b08cbe3-b052-4a50-ae7b-da07c66e348c/6f74480d-ba86-46f2-8d80-76781deff037/zw57sfo_processed.jpeg)
Transcribed Image Text:**Problem 9:**
To cheer him up, someone hands Mikey (still 27.0 kg) a helium-filled balloon. The scale now reads 25.0 kg. What is the force exerted by the balloon? (Include a free-body diagram.)
**Explanation:**
To solve this problem, we need to calculate the force exerted by the helium-filled balloon, which causes the scale reading to drop from Mikey's weight of 27.0 kg to 25.0 kg.
**Steps:**
1. Determine the difference in scale reading:
\[
\text{Difference in weight} = 27.0 \, \text{kg} - 25.0 \, \text{kg} = 2.0 \, \text{kg}
\]
2. Calculate the force exerted by the balloon (which is the buoyant force reducing Mikey’s apparent weight):
\[
F_{\text{balloon}} = \text{Difference in weight} \times g
\]
where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
3. \[
F_{\text{balloon}} = 2.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N}
\]
**Free-Body Diagram:**
The free-body diagram would show:
- Downward force:
- Mikey’s weight: \( 27.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 264.6 \, \text{N} \)
- Upward force:
- Buoyant force by balloon: \( 19.6 \, \text{N} \)
In the diagram, these forces balance out to reflect the apparent weight of 25.0 kg on the scale:
- Apparent weight: \( 25.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 245.0 \, \text{N} \)
This illustrates the decrease in the weight reading due to the upward buoyant force exerted by the helium balloon.
Expert Solution
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Step 1
Given:
mass of Mikey,
Reading of scale after a balloon is held by Mikey,
Step by stepSolved in 2 steps with 1 images
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