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JEE Main 2021 · Physics · Kinematics and Laws of Motion · medium

A block is released from rest at the top of a rough incline of angle 37^° (sin 37^°=0.6) and slides 5 m along it; the coefficient of kinetic friction…

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A block is released from rest at the top of a rough incline of angle 37∘37^\circ (sin⁡37∘=0.6\sin37^\circ=0.6) and slides 5 m5\ \mathrm{m} along it; the coefficient of kinetic friction on the incline is 0.250.25. At the bottom it moves smoothly onto a horizontal surface where the coefficient of kinetic friction is 0.50.5. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the distance it travels on the horizontal surface before stopping is
  1. A. 2 m2\ \mathrm{m}
  2. B. 5 m5\ \mathrm{m}
  3. C. 4 m4\ \mathrm{m}
  4. D. 8 m8\ \mathrm{m}
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Answer: (C) 4 m4\ \mathrm{m}

On the incline a=g(sin⁡θ−μcos⁡θ)=10(0.6−0.25×0.8)=4 m s−2a=g(\sin\theta-\mu\cos\theta)=10(0.6-0.25\times0.8)=4\ \mathrm{m\,s^{-2}}, so v2=2aL=2(4)(5)=40 m2s−2v^2=2aL=2(4)(5)=40\ \mathrm{m^2s^{-2}}. On the horizontal surface the deceleration is μg=5 m s−2\mu g=5\ \mathrm{m\,s^{-2}}, so d=v2/(2μg)=40/10=4 md=v^2/(2\mu g)=40/10=4\ \mathrm{m}. (Work-energy shortcut: mg h−μ1mgcos⁡θ L=μ2mg dmg\,h-\mu_1 mg\cos\theta\,L=\mu_2 mg\,d.)
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