A 0.13 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 300 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The maximum elastic potential energy of the system is closest to: O 0.96 J O 1.7 J O 0.69 J 1.3 J O 1.2 J

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Chapter1: Units, Trigonometry. And Vectors
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Q11

**Problem Statement:**

A 0.13 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 300 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The maximum elastic potential energy of the system is closest to:

- ○ 0.96 J
- ○ 1.7 J
- ○ 0.69 J
- ○ 1.3 J
- ○ 1.2 J

**Explanation:**

The problem involves a block undergoing simple harmonic motion on a frictionless surface. It is attached to a spring, and we are interested in finding out the maximum elastic potential energy stored in the system.

To calculate the maximum elastic potential energy (\(U\)), use the formula:
\[
U = \frac{1}{2} k x^2
\]
where:
- \(k = 300 \, \text{N/m}\) is the force constant of the spring,
- \(x = 0.080 \, \text{m}\) is the displacement from the equilibrium position.

Plug in the values to find \(U\).
Transcribed Image Text:**Problem Statement:** A 0.13 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 300 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The maximum elastic potential energy of the system is closest to: - ○ 0.96 J - ○ 1.7 J - ○ 0.69 J - ○ 1.3 J - ○ 1.2 J **Explanation:** The problem involves a block undergoing simple harmonic motion on a frictionless surface. It is attached to a spring, and we are interested in finding out the maximum elastic potential energy stored in the system. To calculate the maximum elastic potential energy (\(U\)), use the formula: \[ U = \frac{1}{2} k x^2 \] where: - \(k = 300 \, \text{N/m}\) is the force constant of the spring, - \(x = 0.080 \, \text{m}\) is the displacement from the equilibrium position. Plug in the values to find \(U\).
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