An object is moving up and down on a spring (mass-spring system) on Earth. The object has a mass of 1.5 kg and the spring constant of the spring is 50 N/m. What is the period of the object's motion? Your answer: 0.03 s 1.1 s 2.5 s O 14 s
An object is moving up and down on a spring (mass-spring system) on Earth. The object has a mass of 1.5 kg and the spring constant of the spring is 50 N/m. What is the period of the object's motion? Your answer: 0.03 s 1.1 s 2.5 s O 14 s
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![**Mass-Spring System Problem**
**Problem Statement:**
An object is moving up and down on a spring (mass-spring system) on Earth. The object has a mass \( \mathbf{m} \) of 1.5 kg and the spring constant \( \mathbf{k} \) of the spring is 50 N/m. What is the period of the object’s motion?
**Answer Choices:**
- [ ] 0.03 s
- [ ] 1.1 s
- [ ] 2.5 s
- [ ] 14 s
### Explanation:
The period \( \mathbf{T} \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Where:
- \( \mathbf{m} \) is the mass of the object.
- \( \mathbf{k} \) is the spring constant.
- \( \mathbf{T} \) is the period of the motion.
### Steps to Solve:
1. Identify the given values:
- Mass, \( m = 1.5 \) kg
- Spring constant, \( k = 50 \) N/m
2. Plug these values into the formula:
\[ T = 2\pi \sqrt{\frac{1.5\, \text{kg}}{50\, \text{N/m}}} \]
3. Perform the calculation inside the square root:
\[ \frac{1.5\, \text{kg}}{50\, \text{N/m}} = 0.03\, \text{kg/N} \]
4. Take the square root:
\[ \sqrt{0.03\, \text{kg/N}} \approx 0.1732 \, \text{s} \]
5. Multiply by \( 2\pi \):
\[ T \approx 2\pi \times 0.1732 \, \text{s} \approx 1.088 \, \text{s} \]
6. Round to the nearest tenths place if necessary:
\[ T \approx 1.1 \, \text{s} \]
Based on the calculation, the correct answer is **1.1 s**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc53b197f-77a2-4852-8dbf-ece724a1e802%2Fcdca294a-d4f2-4211-89cc-3af3c5c76ce8%2Ff6cbtkk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mass-Spring System Problem**
**Problem Statement:**
An object is moving up and down on a spring (mass-spring system) on Earth. The object has a mass \( \mathbf{m} \) of 1.5 kg and the spring constant \( \mathbf{k} \) of the spring is 50 N/m. What is the period of the object’s motion?
**Answer Choices:**
- [ ] 0.03 s
- [ ] 1.1 s
- [ ] 2.5 s
- [ ] 14 s
### Explanation:
The period \( \mathbf{T} \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Where:
- \( \mathbf{m} \) is the mass of the object.
- \( \mathbf{k} \) is the spring constant.
- \( \mathbf{T} \) is the period of the motion.
### Steps to Solve:
1. Identify the given values:
- Mass, \( m = 1.5 \) kg
- Spring constant, \( k = 50 \) N/m
2. Plug these values into the formula:
\[ T = 2\pi \sqrt{\frac{1.5\, \text{kg}}{50\, \text{N/m}}} \]
3. Perform the calculation inside the square root:
\[ \frac{1.5\, \text{kg}}{50\, \text{N/m}} = 0.03\, \text{kg/N} \]
4. Take the square root:
\[ \sqrt{0.03\, \text{kg/N}} \approx 0.1732 \, \text{s} \]
5. Multiply by \( 2\pi \):
\[ T \approx 2\pi \times 0.1732 \, \text{s} \approx 1.088 \, \text{s} \]
6. Round to the nearest tenths place if necessary:
\[ T \approx 1.1 \, \text{s} \]
Based on the calculation, the correct answer is **1.1 s**.
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