A weight of 10 kg is hung is fixed to the ceiling and is 1 meter above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is
Correct answer: B. 0.05 joule
- A. Zero
- B. 0.05 joule
- C. 100 joule
- D. 500 joule
Explanation
The energy stored in a stretched wire is given by the formula U = (1/2) × F × e, where F is the force applied, and e is the elongation. Here, the force F is the weight of the object, which is 10 kg × 9.8 m/s2 = 98 N. The elongation e is 1 mm, which is 0.001 m. Substituting these values, the energy U = 0.5 × 98 N × 0.001 m = 0.049 J, which rounds to 0.05 J. Therefore, option B is correct. Other options are incorrect as they either suggest no energy storage or provide unrealistic values of energy.
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