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JEE Main 2016 · Physics · Kinematics and Laws of Motion · hard

A block of mass 2 kg rests at one end of a 3 m long plank of mass 4 kg lying on a smooth horizontal floor. The coefficient of friction between block…

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A block of mass 2 kg2\ \mathrm{kg} rests at one end of a 3 m3\ \mathrm{m} long plank of mass 4 kg4\ \mathrm{kg} lying on a smooth horizontal floor. The coefficient of friction between block and plank is 0.30.3. A horizontal force of 15 N15\ \mathrm{N} is applied to the block along the plank. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. The block leaves the other end of the plank after a time
  1. A. 1 s1\ \mathrm{s}
  2. B. 2 s\sqrt{2}\ \mathrm{s}
  3. C. 3 s\sqrt{3}\ \mathrm{s}
  4. D. 2 s2\ \mathrm{s}
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Answer: (B) 2 s\sqrt{2}\ \mathrm{s}

First test for slipping. The largest acceleration friction can give the plank is μmg/M=0.3×2×10/4=1.5 m s−2\mu m g/M=0.3\times2\times10/4=1.5\ \mathrm{m\,s^{-2}}. If both moved together the acceleration would be 15/6=2.5 m s−2>1.515/6=2.5\ \mathrm{m\,s^{-2}}>1.5, so the block slides. Block: a1=(15−6)/2=4.5 m s−2a_1=(15-6)/2=4.5\ \mathrm{m\,s^{-2}}. Plank: a2=6/4=1.5 m s−2a_2=6/4=1.5\ \mathrm{m\,s^{-2}}. The relative acceleration is 3 m s−23\ \mathrm{m\,s^{-2}} and the relative displacement must be the plank length: 3=12(3)t23=\tfrac12(3)t^2, so t2=2t^2=2 and t=2 st=\sqrt2\ \mathrm{s}. Using a=15/6a=15/6 without checking for slipping would give a wrong answer.
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