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NEET 2020 (Phase 2) · Friction Basics

Friction Basics question from NEET 2020 (Phase 2)

Q21NEET 2020 · PHASE 2Medium

Friction Basics

Calculate the acceleration of the block and trolly system shown in the figure. The coefficient of kinetic friction between the trolly and the surface is 0.05. (g=10 m/s2g = 10\text{ m/s}^{2}, mass of the string is negligible and no other friction exists).

Figure

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Answer: (b)

  1. Driving force: weight of block =2×10=20 N= 2 \times 10 = 20\text{ N}.
  2. Friction on trolley =μmg=0.05×10×10=5 N= \mu m g = 0.05 \times 10 \times 10 = 5\text{ N}.
  3. a=20−510+2=1512a = \dfrac{20 - 5}{10 + 2} = \dfrac{15}{12}
  4. So the answer is (b) 1.25 m/s21.25\text{ m/s}^2.
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