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JEE Main 2024 · Physics · Work, Energy and Power · medium

A 2 kg block is released from rest on a smooth incline of 30^°. After sliding 1.5 m along the incline it hits a light spring of force constant 200 N…

Single correct option

A 2 kg2\ \mathrm{kg} block is released from rest on a smooth incline of 30∘30^\circ. After sliding 1.5 m1.5\ \mathrm{m} along the incline it hits a light spring of force constant 200 N m−1200\ \mathrm{N\,m^{-1}} lying along the incline. The maximum compression of the spring is nearly (g=10 m s−2g=10\ \mathrm{m\,s^{-2}})
  1. A. 0.44 m0.44\ \mathrm{m}
  2. B. 0.30 m0.30\ \mathrm{m}
  3. C. 0.37 m0.37\ \mathrm{m}
  4. D. 0.55 m0.55\ \mathrm{m}
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Answer: (A) 0.44 m0.44\ \mathrm{m}

Energy conservation from release to maximum compression xx: the block descends a total slant distance (1.5+x)(1.5+x), losing height (1.5+x)sin⁡30∘(1.5+x)\sin30^\circ. So mg(1.5+x)sin⁡30∘=12kx2mg(1.5+x)\sin30^\circ=\tfrac12kx^2, i.e. 10(1.5+x)=100x210(1.5+x)=100x^2, giving 20x2−2x−3=020x^2-2x-3=0 and x=2+24440≈0.44 mx=\dfrac{2+\sqrt{244}}{40}\approx0.44\ \mathrm{m}. Ignoring the extra descent during compression gives 0.390.39 m, a trap.
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