Let's break this down step-by-step.

Problem Overview:

• Spring constant (k): 17 N/m
• Relaxed length of spring (L₀): 0.23 m
• Mass of the block (m): 0.090 kg
• Initial stretched length of spring (L₁): 0.29 m
• Gravity (g): 10 m/s²
• Time (t): 0.06 s
• Origin (y = 0): Top of the shelf, with the +y direction being upward.

Steps to Solve:

1. Determine the force at the equilibrium position: When the block is at rest after being stretched, the force on the block is the spring force and the gravitational force. At equilibrium, these forces balance.

   The spring force is given by Hooke's Law: $F_{\text{spring}} = k \Delta y$ where $\Delta y = L_1 - L_0 = 0.29 \, \text{m} - 0.23 \, \text{m} = 0.06 \, \text{m}$.

   So, $F_{\text{spring}} = 17 \times 0.06 = 1.02 \, \text{N}$ The gravitational force is: $F_{\text{gravity}} = m \cdot g = 0.090 \, \text{kg} \times 10 \, \text{m/s}^2 = 0.90 \, \text{N}$

   The spring force is upwards, and gravity is downwards. The net force at equilibrium is zero, as the spring force exactly balances gravity.

2. Set up the equation for oscillation: After the release, the block will undergo simple harmonic motion (SHM) about the equilibrium position. The displacement $y(t)$ from the equilibrium position follows the SHM equation:

   $y(t) = A \cos(\omega t + \phi)$ where:

   • $A$ is the amplitude of oscillation,
   • $\omega$ is the angular frequency,
   • $\phi$ is the phase constant.

   To find $A$, we note that the block was initially pulled to a length of 0.29 m, and when released, it oscillates from this position. Therefore, the amplitude is the initial displacement from equilibrium: $A = 0.29 \, \text{m} - 0.23 \, \text{m} = 0.06 \, \text{m}$

3. Find the angular frequency $\omega$: The angular frequency for SHM is: $\omega = \sqrt{\frac{k}{m}}$ Substituting values: $\omega = \sqrt{\frac{17}{0.090}} \approx \sqrt{188.89} \approx 13.75 \, \text{rad/s}$

4. Determine the phase constant $\phi$: Since the block is released from rest at the maximum displacement, the cosine function starts at its maximum value, so $\phi = 0$.

5. Find the displacement at $t = 0.06$ s: Using the equation for SHM: $y(t) = A \cos(\omega t)$ Substituting the known values: $y(0.06) = 0.06 \cos(13.75 \times 0.06)$ $y(0.06) = 0.06 \cos(0.825)$ Using $\cos(0.825) \approx 0.670$: $y(0.06) \approx 0.06 \times 0.670 = 0.0402 \, \text{m}$

Thus, after 0.06 seconds, the displacement of the block relative to the shelf is approximately 0.0402 m above the equilibrium position (so above the shelf).

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Final Answer:

The $y$-component of the block’s position relative to the shelf at $t = 0.06$ s is approximately 0.0402 m.

Would you like a more detailed breakdown of any step, or do you have any questions about the calculation?

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Related Questions:

1. How does the spring constant $k$ affect the oscillation period?
2. What happens to the amplitude of oscillation if the spring is stretched more or less?
3. How would the frequency change if the block had a different mass?
4. How would the phase change if the block were released from a different position?
5. What is the period of the oscillation of this system?

Tip:

When dealing with oscillations, always keep track of the equilibrium position (where forces are balanced) and use it as a reference point for your calculations!