10. A vertically mounted spring (k = 750 N/m) is compressed by 30.0 cm relative to its unstrained length. A mass (m = 0.28 kg) is placed at rest against the spring. When the spring is released, the mass is launched vertically in the air. What is the speed of the mass (m) when it passes by a point located 4.10 m above the release point? m/s

Physics for Scientists and Engineers: Foundations and Connections
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Chapter8: Conservation Of Energy
Section: Chapter Questions
Problem 24PQ: A block is placed on top of a vertical spring, and the spring compresses. Figure P8.24 depicts a...
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**Question 10:**

A vertically mounted spring (k = 750 N/m) is compressed by 30.0 cm relative to its unstrained length. A mass (m = 0.28 kg) is placed at rest against the spring. When the spring is released, the mass is launched vertically in the air. What is the speed of the mass (m) when it passes by a point located 4.10 m above the release point?

\[ \_\_\_\_\_\_ \text{ m/s} \]

---

**Explanation:**

This problem involves concepts from Hooke's Law and the principles of energy conservation. Let's analyze each part of the question to understand the process you might go through to solve it.

1. **Spring Compression:**
   - The spring constant \( k \) is given as 750 N/m.
   - The spring is compressed by 30.0 cm, which is 0.30 m.
   - The energy stored in the compressed spring is calculated using the formula for elastic potential energy:
   
   \[
   E_{\text{spring}} = \frac{1}{2} k x^2
   \]
   
   Here, \( x \) is the compression distance.

2. **Mass and Gravitational Potential Energy:**
   - The mass \( m \) is 0.28 kg.
   - When the spring releases the mass, it converts the elastic potential energy into kinetic energy and eventually potential energy as it rises.
   - To find the speed at 4.10 m above the release point, consider the conservation of energy.
   
   The total mechanical energy when the spring is released (initially all from the spring) will be distributed between kinetic energy and gravitational potential energy at the point of interest (4.10 m above).

3. **Gravitational Potential Energy Change:**
   - The gravitational potential energy change \( \Delta E_{\text{gravity}} \) when the mass moves a height \( h \) (4.10 m in this case) can be calculated by:

   \[
   E_{\text{gravity}} = mgh
   \]
   
   Here, \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).

Using these formulas, you can calculate the speed of the mass at the given height.

---

**Calculation Steps:**

1. Calculate
Transcribed Image Text:**Question 10:** A vertically mounted spring (k = 750 N/m) is compressed by 30.0 cm relative to its unstrained length. A mass (m = 0.28 kg) is placed at rest against the spring. When the spring is released, the mass is launched vertically in the air. What is the speed of the mass (m) when it passes by a point located 4.10 m above the release point? \[ \_\_\_\_\_\_ \text{ m/s} \] --- **Explanation:** This problem involves concepts from Hooke's Law and the principles of energy conservation. Let's analyze each part of the question to understand the process you might go through to solve it. 1. **Spring Compression:** - The spring constant \( k \) is given as 750 N/m. - The spring is compressed by 30.0 cm, which is 0.30 m. - The energy stored in the compressed spring is calculated using the formula for elastic potential energy: \[ E_{\text{spring}} = \frac{1}{2} k x^2 \] Here, \( x \) is the compression distance. 2. **Mass and Gravitational Potential Energy:** - The mass \( m \) is 0.28 kg. - When the spring releases the mass, it converts the elastic potential energy into kinetic energy and eventually potential energy as it rises. - To find the speed at 4.10 m above the release point, consider the conservation of energy. The total mechanical energy when the spring is released (initially all from the spring) will be distributed between kinetic energy and gravitational potential energy at the point of interest (4.10 m above). 3. **Gravitational Potential Energy Change:** - The gravitational potential energy change \( \Delta E_{\text{gravity}} \) when the mass moves a height \( h \) (4.10 m in this case) can be calculated by: \[ E_{\text{gravity}} = mgh \] Here, \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)). Using these formulas, you can calculate the speed of the mass at the given height. --- **Calculation Steps:** 1. Calculate
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