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JEE Main 2017 · Physics · Work, Energy and Power · medium

A block of mass 2 kg is released from rest from a height of 0.6 m above the free end of a vertical light spring of force constant 400 N m^-1 fixed on…

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A block of mass 2 kg2\ \mathrm{kg} is released from rest from a height of 0.6 m0.6\ \mathrm{m} above the free end of a vertical light spring of force constant 400 N m−1400\ \mathrm{N\,m^{-1}} fixed on the ground. Taking g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}, the maximum compression of the spring is
  1. A. 0.30 m0.30\ \mathrm{m}
  2. B. 0.25 m0.25\ \mathrm{m}
  3. C. 0.35 m0.35\ \mathrm{m}
  4. D. 0.15 m0.15\ \mathrm{m}
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Answer: (A) 0.30 m0.30\ \mathrm{m}

Take the lowest point as the final state, where the block is momentarily at rest. The block falls a total height 0.6+x0.6 + x, and all this gravitational energy goes into the spring: mg(h+x)=12kx2mg(h + x) = \tfrac12kx^2. So 20(0.6+x)=200x220(0.6 + x) = 200x^2, i.e. 200x2−20x−12=0200x^2 - 20x - 12 = 0, or 50x2−5x−3=050x^2 - 5x - 3 = 0. This gives x=5+25+600100=5+25100=0.30 mx = \dfrac{5 + \sqrt{25 + 600}}{100} = \dfrac{5 + 25}{100} = 0.30\ \mathrm{m}. Ignoring the further fall during compression, mgh=12kx2mgh = \tfrac12kx^2, gives 0.245 m0.245\ \mathrm{m}, the nearest distractor.
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