(a) The figure below shows an 8.00 kg stone at rest on a spring. The spring is compressed 10.0 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravi- tational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point? 00000000000

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### Educational Exercise: Analyzing a Stone on a Spring

#### Problem Statement:

The figure below shows an 8.00 kg stone at rest on a spring. The spring is compressed 10.0 cm by the stone.

(a) What is the spring constant?

(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?

(c) What is the change in the gravitational potential energy of the stone–Earth system when the stone moves from the release point to its maximum height?

(d) What is that maximum height, measured from the release point?

#### Diagram Description:

The figure depicts an 8.00 kg stone resting on a vertically aligned spring, which is anchored on a flat surface. The spring is compressed under the weight of the stone, with a spring constant denoted by \( k \).

![Spring Diagram](insert image URL here)

#### Detailed Explanation:

1. **Spring Constant Determination (a)**:
    - The spring constant \( k \) can be determined by applying Hooke's Law in the context of the force exerted by the stone due to gravity. The formula is \( F = k \cdot x \), where \( F \) is the force applied by the weight of the stone and \( x \) is the compression distance.
    - Given: 
      \[ F = m \cdot g \] 
      \[ x = 10.0 \, \text{cm} = 0.1 \, \text{m} \] 
      where \( m = 8.00 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
    - Calculation for \( k \):
      \[
      k = \frac{F}{x} = \frac{(8.00 \, \text{kg}) (9.8 \, \text{m/s}^2)}{0.1 \, \text{m}}
      \]

2. **Elastic Potential Energy Calculation (b)**:
    - When the stone is pushed further down by 30.0 cm (0.3 m):
      The total compression \( x_{\text{total}} = 0.1 \, \text{m} + 0.3 \, \text{m} = 0.
Transcribed Image Text:### Educational Exercise: Analyzing a Stone on a Spring #### Problem Statement: The figure below shows an 8.00 kg stone at rest on a spring. The spring is compressed 10.0 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stone–Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point? #### Diagram Description: The figure depicts an 8.00 kg stone resting on a vertically aligned spring, which is anchored on a flat surface. The spring is compressed under the weight of the stone, with a spring constant denoted by \( k \). ![Spring Diagram](insert image URL here) #### Detailed Explanation: 1. **Spring Constant Determination (a)**: - The spring constant \( k \) can be determined by applying Hooke's Law in the context of the force exerted by the stone due to gravity. The formula is \( F = k \cdot x \), where \( F \) is the force applied by the weight of the stone and \( x \) is the compression distance. - Given: \[ F = m \cdot g \] \[ x = 10.0 \, \text{cm} = 0.1 \, \text{m} \] where \( m = 8.00 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). - Calculation for \( k \): \[ k = \frac{F}{x} = \frac{(8.00 \, \text{kg}) (9.8 \, \text{m/s}^2)}{0.1 \, \text{m}} \] 2. **Elastic Potential Energy Calculation (b)**: - When the stone is pushed further down by 30.0 cm (0.3 m): The total compression \( x_{\text{total}} = 0.1 \, \text{m} + 0.3 \, \text{m} = 0.
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