Energy Conversion In A Falling Body
A stone of mass 0.5 kg is dropped from rest at a height of 8 m above the ground. Ignoring air resistance and taking g as 10 m/s squared, what is its kinetic energy just before it strikes the ground?
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About This Question
- Subject
- physics
- Chapter
- work, energy and power
- Topic
- energy conversion in a falling body
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
40 J
As described in NCERT Class 11, Chapter 6 (Work, Energy and Power), for a freely falling body with no air resistance, mechanical energy is conserved, so the kinetic energy gained equals the gravitational potential energy lost. The initial potential energy is mgh = 0.5 × 10 × 8 = 40 J. Since the stone starts from rest, all of this converts into kinetic energy just before impact, giving KE = 40 J. The option 20 J halves the result, perhaps by misapplying a factor of one-half meant for kinetic energy. The option 80 J doubles the height or mass erroneously. The option 4 J misplaces a decimal in the mass. A magnitude check using v = sqrt(2gh) = sqrt(160) gives about 12.6 m/s, and (1/2)(0.5)(160) = 40 J, confirming the kinetic energy independently of the potential-energy route.
This easy difficulty physics question is from the chapter work, energy and power, covering the topic of energy conversion in a falling body. It appeared in the 2025 exam.
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