A 22 kq object is mass is pulled 0.35m downward from its equilibrium position and allowed to oscillate. What is the maxium Kinetic enerau or the Suspended from a spring with k=18 Nim. The Object?

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Chapter1: Units, Trigonometry. And Vectors
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### Problem Statement
A 2.2 kg object is suspended from a spring with k = 18 N/m. The mass is pulled 0.35 m downward from its equilibrium position and allowed to oscillate. What is the maximum kinetic energy of the object?

### Explanation:
In this scenario, we can determine the maximum kinetic energy of the object using the law of conservation of energy. When the object is at its maximum displacement (0.35 m), the entire energy of the system is in the form of elastic potential energy stored in the spring. As the object passes through the equilibrium position, this potential energy is converted into kinetic energy.

The elastic potential energy (U) in a spring is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where:
- \( k \) is the spring constant (18 N/m),
- \( x \) is the displacement from equilibrium (0.35 m).

Plugging in these values:
\[ U = \frac{1}{2} \times 18 \, \text{N/m} \times (0.35 \, \text{m})^2 \]
\[ U = \frac{1}{2} \times 18 \times 0.1225 \]
\[ U = 1.1025 \, \text{J} \]

Therefore, the maximum kinetic energy of the object is 1.1025 Joules.
Transcribed Image Text:### Problem Statement A 2.2 kg object is suspended from a spring with k = 18 N/m. The mass is pulled 0.35 m downward from its equilibrium position and allowed to oscillate. What is the maximum kinetic energy of the object? ### Explanation: In this scenario, we can determine the maximum kinetic energy of the object using the law of conservation of energy. When the object is at its maximum displacement (0.35 m), the entire energy of the system is in the form of elastic potential energy stored in the spring. As the object passes through the equilibrium position, this potential energy is converted into kinetic energy. The elastic potential energy (U) in a spring is given by the formula: \[ U = \frac{1}{2} k x^2 \] where: - \( k \) is the spring constant (18 N/m), - \( x \) is the displacement from equilibrium (0.35 m). Plugging in these values: \[ U = \frac{1}{2} \times 18 \, \text{N/m} \times (0.35 \, \text{m})^2 \] \[ U = \frac{1}{2} \times 18 \times 0.1225 \] \[ U = 1.1025 \, \text{J} \] Therefore, the maximum kinetic energy of the object is 1.1025 Joules.
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