10. An object of mass 0.55 kg is attached to a spring, with spring constant 11.0 N/m. If the object is stretched 4.50 cm from the equilibrium position and released, what is its maximum speed? m/s sf6 60 s Jss €60 ssfe

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Chapter13: Vibrations And Waves
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Problem 14P: An object-spring system moving with simple harmonic motion has an amplitude A. (a) What is the total...
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### Physics Problem: Determining Maximum Speed of an Oscillating Object

**Problem Statement:**
10. An object of mass 0.55 kg is attached to a spring, with a spring constant 11.0 N/m. If the object is stretched 4.50 cm from the equilibrium position and released, what is its maximum speed?

[_____] m/s

### Explanation:

In this problem, we are dealing with an object undergoing simple harmonic motion. We need to determine the maximum speed of the object.

**Given:**
- Mass of the object, \( m = 0.55 \, \text{kg} \)
- Spring constant, \( k = 11.0 \, \text{N/m} \)
- Displacement from equilibrium, \( x = 4.50 \, \text{cm} = 0.045 \, \text{m} \)

**Concepts Involved:**
- Hooke's Law: \( F = -kx \)
- Conservation of Energy in Simple Harmonic Motion:
  \[
  \frac{1}{2} k x^2 = \frac{1}{2} m v_{\text{max}}^2
  \]
  Where:
  - \( v_{\text{max}} \) is the maximum speed of the object.

**Steps to Solve:**
1. Using the conservation of mechanical energy, set the potential energy at the maximum displacement equal to the kinetic energy at the equilibrium position.
   \[
   \frac{1}{2} k x^2 = \frac{1}{2} m v_{\text{max}}^2
   \]
   
2. Simplify and solve for \( v_{\text{max}} \):
   \[
   k x^2 = m v_{\text{max}}^2
   \]
   \[
   v_{\text{max}} = \sqrt{\frac{k x^2}{m}}
   \]
   
3. Substitute the given values into the equation:
   \[
   v_{\text{max}} = \sqrt{\frac{11.0 \times (0.045)^2}{0.55}}
   \]
   
4. Calculate the result:
   \[
   v_{\text{max}} = \sqrt{\frac{11.0 \times 0.002025}{0.55}} = \sqrt{\frac{0.
Transcribed Image Text:### Physics Problem: Determining Maximum Speed of an Oscillating Object **Problem Statement:** 10. An object of mass 0.55 kg is attached to a spring, with a spring constant 11.0 N/m. If the object is stretched 4.50 cm from the equilibrium position and released, what is its maximum speed? [_____] m/s ### Explanation: In this problem, we are dealing with an object undergoing simple harmonic motion. We need to determine the maximum speed of the object. **Given:** - Mass of the object, \( m = 0.55 \, \text{kg} \) - Spring constant, \( k = 11.0 \, \text{N/m} \) - Displacement from equilibrium, \( x = 4.50 \, \text{cm} = 0.045 \, \text{m} \) **Concepts Involved:** - Hooke's Law: \( F = -kx \) - Conservation of Energy in Simple Harmonic Motion: \[ \frac{1}{2} k x^2 = \frac{1}{2} m v_{\text{max}}^2 \] Where: - \( v_{\text{max}} \) is the maximum speed of the object. **Steps to Solve:** 1. Using the conservation of mechanical energy, set the potential energy at the maximum displacement equal to the kinetic energy at the equilibrium position. \[ \frac{1}{2} k x^2 = \frac{1}{2} m v_{\text{max}}^2 \] 2. Simplify and solve for \( v_{\text{max}} \): \[ k x^2 = m v_{\text{max}}^2 \] \[ v_{\text{max}} = \sqrt{\frac{k x^2}{m}} \] 3. Substitute the given values into the equation: \[ v_{\text{max}} = \sqrt{\frac{11.0 \times (0.045)^2}{0.55}} \] 4. Calculate the result: \[ v_{\text{max}} = \sqrt{\frac{11.0 \times 0.002025}{0.55}} = \sqrt{\frac{0.
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