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JEE Main 2007 (AIEEE) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2007 (AIEEE)

Q27JEE Main 2007 · AIEEEMedium

Spring & Elastic Energy

A 2 kg2\text{ kg} block slides on a horizontal floor with a speed of 4 m/s.4\text{ m/s}. It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N15\text{ N} and spring constant is 10,000 N/m.10{,}000\text{ N/m}. The spring compresses by

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Answer: (b)

  1. Work-energy: 12mv2=fx+12kx2\frac{1}{2}mv^2 = fx + \frac{1}{2}kx^2.
  2. 16=15x+5000x2⇒5000x2+15x−16=016 = 15x + 5000x^2 \Rightarrow 5000x^2 + 15x - 16 = 0
  3. x=−15+225+32000010000≈0.055x = \frac{-15 + \sqrt{225 + 320000}}{10000} \approx 0.055 m
  4. So the answer is (b) 5.55.5 cm.
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