An elevator is moving upward at 5 m/s while slowing down at 2 m/s/s. Inside the elevator a yellow box (5.0 kg) is stacked on top of a gray box (10.0 kg). 1. If a scale is placed between the boxes, what will the scale read? 2. If a scale is placed beneath the bottom box, what will the scale read? 37

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### Student Whiteboard Problem

**Educational Focus: Understanding Forces in an Accelerating Frame**

---

**Problem Statement:**

An elevator is moving upward at 5 m/s while slowing down at 2 m/s². Inside the elevator, a yellow box (5.0 kg) is stacked on top of a gray box (10.0 kg).

#### Questions:
1. If a scale is placed between the boxes, what will the scale read?
2. If a scale is placed beneath the bottom box, what will the scale read?

---

**Diagram Description:**

A diagram is provided that shows a simple representation of an elevator. Inside the elevator, a yellow box is stacked on top of a gray box. 

**Detailed Explanation of the Diagram:**

- The elevator is shown as a vertical rectangle with a cable attached to its top, indicating it is suspended.
- Inside the elevator, a small yellow box (representing the 5.0 kg box) is stacked directly on top of a larger gray box (representing the 10.0 kg box).

---

**Considerations for Solving the Problem:**

- Since the elevator is moving upward but slowing down, it is experiencing a downward acceleration of 2 m/s².
- The effective gravitational force acting on the boxes will be modified by the elevator's acceleration.

**Step-by-Step Solution:**
1. **For the scale between the boxes:**

    - Calculate the effective gravitational acceleration: 
      \( g_{eff} = g - a = 9.8 \, m/s² - 2.0 \, m/s² = 7.8 \, m/s² \).
    - Determine the force on the scale due to the top box:
      \( F_{scale} = m_{yellow} \times g_{eff} = 5.0 \, kg \times 7.8 \, m/s² = 39 \, N \).

2. **For the scale beneath the bottom box:**

    - The scale must support the entire weight of both boxes.
    - Total mass:
      \( m_{total} = m_{yellow} + m_{gray} = 5.0 \, kg + 10.0 \, kg = 15.0 \, kg \).
    - Calculate the force exerted on the scale:
      \( F_{total} = m_{total} \times g_{eff} = 15.
Transcribed Image Text:### Student Whiteboard Problem **Educational Focus: Understanding Forces in an Accelerating Frame** --- **Problem Statement:** An elevator is moving upward at 5 m/s while slowing down at 2 m/s². Inside the elevator, a yellow box (5.0 kg) is stacked on top of a gray box (10.0 kg). #### Questions: 1. If a scale is placed between the boxes, what will the scale read? 2. If a scale is placed beneath the bottom box, what will the scale read? --- **Diagram Description:** A diagram is provided that shows a simple representation of an elevator. Inside the elevator, a yellow box is stacked on top of a gray box. **Detailed Explanation of the Diagram:** - The elevator is shown as a vertical rectangle with a cable attached to its top, indicating it is suspended. - Inside the elevator, a small yellow box (representing the 5.0 kg box) is stacked directly on top of a larger gray box (representing the 10.0 kg box). --- **Considerations for Solving the Problem:** - Since the elevator is moving upward but slowing down, it is experiencing a downward acceleration of 2 m/s². - The effective gravitational force acting on the boxes will be modified by the elevator's acceleration. **Step-by-Step Solution:** 1. **For the scale between the boxes:** - Calculate the effective gravitational acceleration: \( g_{eff} = g - a = 9.8 \, m/s² - 2.0 \, m/s² = 7.8 \, m/s² \). - Determine the force on the scale due to the top box: \( F_{scale} = m_{yellow} \times g_{eff} = 5.0 \, kg \times 7.8 \, m/s² = 39 \, N \). 2. **For the scale beneath the bottom box:** - The scale must support the entire weight of both boxes. - Total mass: \( m_{total} = m_{yellow} + m_{gray} = 5.0 \, kg + 10.0 \, kg = 15.0 \, kg \). - Calculate the force exerted on the scale: \( F_{total} = m_{total} \times g_{eff} = 15.
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