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NEET 2009 (AIPMT) · Work-Energy Theorem

Work-Energy Theorem question from NEET 2009 (AIPMT)

Q25NEET 2009 · AIPMTEasy

Work-Energy Theorem

A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s2)

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Answer: (d)

  1. Initial KE =12(1)(20)2=200 J= \frac{1}{2}(1)(20)^2 = 200\text{ J}
  2. PE gained at 18 m =mgh=1×10×18=180 J= mgh = 1 \times 10 \times 18 = 180\text{ J}
  3. Energy lost to air friction =200−180=20 J= 200 - 180 = 20\text{ J}, so the answer is (d).
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