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.)

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Chapter1: Units, Trigonometry. And Vectors
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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.
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
Step 1

Given:

mass of Mikey, m=27 kg

Reading of scale after a balloon is held by Mikey, m'=25 kg

 

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