Elastic Potential Energy
Spring of k = 200 N/m compressed by 0.1 m stores energy:
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Solution
1 J
The elastic potential energy (U) stored in a spring compressed or stretched by distance x from its equilibrium position is: U=21kx2 Given:
- k=200 N/m
- x=0.1 m U=21(200)(0.1)2=100×0.01=1 Joule.
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About This Question
- Subject
- physics
- Chapter
- work, energy and power
- Topic
- elastic potential energy
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1 J
The elastic potential energy (U) stored in a spring compressed or stretched by distance x from its equilibrium position is: U=21kx2 Given:
- k=200 N/m
- x=0.1 m U=21(200)(0.1)2=100×0.01=1 Joule.
This medium difficulty physics question is from the chapter work, energy and power, covering the topic of elastic potential energy. It appeared in the 2025 exam.
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