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NEET 2002 (AIPMT) · Friction Basics

Friction Basics question from NEET 2002 (AIPMT)

Q36NEET 2002 · AIPMTEasy

Friction Basics

A block of mass 10 kg placed on rough horizontal surface having coefficient of friction μ\mu = 0.5, if a horizontal force of 100 N acting on it then acceleration of the block will be

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Answer: (b)

  1. Friction f=μmg=0.5×10×10=50f = \mu mg = 0.5 \times 10 \times 10 = 50 N (taking g=10 m/s2g = 10\text{ m/s}^2).
  2. a=F−fm=100−5010a = \dfrac{F - f}{m} = \dfrac{100 - 50}{10}
  3. a=5 m/s2a = 5\text{ m/s}^2, so the answer is (b).
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