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JEE Main 2019 · Physics · Work, Energy and Power · medium

A block of mass 2 kg is released from rest on a smooth plane inclined at 30^°. It slides 0.6 m along the plane before touching the free end of a…

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A block of mass 2 kg2\ \mathrm{kg} is released from rest on a smooth plane inclined at 30∘30^\circ. It slides 0.6 m0.6\ \mathrm{m} along the plane before touching the free end of a light spring of force constant 200 N m−1200\ \mathrm{N\,m^{-1}} lying along the incline, with its other end fixed at the bottom. Take 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. 0.24 m0.24\ \mathrm{m}
  3. C. 0.35 m0.35\ \mathrm{m}
  4. D. 0.60 m0.60\ \mathrm{m}
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Answer: (A) 0.30 m0.30\ \mathrm{m}

The block keeps sliding while the spring compresses, so the total distance descended along the plane is 0.6+x0.6 + x. Energy conservation (block at rest at start and at maximum compression): mgsin⁡30∘ (0.6+x)=12kx2mg\sin30^\circ\,(0.6+x) = \tfrac12 k x^2. With mgsin⁡30∘=10 Nmg\sin30^\circ = 10\ \mathrm{N} this reads 10(0.6+x)=100x210(0.6+x)=100x^2, i.e. 10x2−x−0.6=010x^2 - x - 0.6 = 0. Hence x=1+1+2420=0.30 mx = \dfrac{1+\sqrt{1+24}}{20} = 0.30\ \mathrm{m}. Ignoring the extra descent during compression gives x=0.06≈0.24 mx=\sqrt{0.06}\approx 0.24\ \mathrm{m}, a typical error.
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