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JEE Main 2020 · Physics · Kinematics and Laws of Motion · medium

A block of mass 2 kg lies on a rough incline of angle 37^° (take sin 37^° = 0.6) with coefficient of friction 0.5. A purely horizontal force F pushes…

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A block of mass 2 kg2\ \mathrm{kg} lies on a rough incline of angle 37∘37^\circ (take sin⁡37∘=0.6\sin 37^\circ = 0.6) with coefficient of friction 0.50.5. A purely horizontal force FF pushes the block towards the incline so that it moves up the slope with constant speed. Taking g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}, the value of FF is
  1. A. 25 N25\ \mathrm{N}
  2. B. 40 N40\ \mathrm{N}
  3. C. 50 N50\ \mathrm{N}
  4. D. 18 N18\ \mathrm{N}
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Answer: (B) 40 N40\ \mathrm{N}

Resolve along and perpendicular to the incline. Normal reaction: N=mgcos⁡θ+Fsin⁡θN = mg\cos\theta + F\sin\theta (the horizontal push has a component into the plane). Along the slope, for zero acceleration: Fcos⁡θ=mgsin⁡θ+μNF\cos\theta = mg\sin\theta + \mu N. So F(cos⁡θ−μsin⁡θ)=mg(sin⁡θ+μcos⁡θ)F(\cos\theta - \mu\sin\theta) = mg(\sin\theta + \mu\cos\theta), i.e. F(0.8−0.3)=20(0.6+0.4)F(0.8 - 0.3) = 20(0.6 + 0.4) giving F=40 NF = 40\ \mathrm{N}. Forgetting the Fsin⁡θF\sin\theta term in NN gives the trap value 25 N25\ \mathrm{N}.
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