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JEE Main 2019 (10 Apr, Shift 2) · Two-Body Friction

Two-Body Friction question from JEE Main 2019 (10 Apr, Shift 2)

Q8JEE Main 2019 · 10 APR · SHIFT 2Medium

Two-Body Friction

Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is:

[Take g = 10 m/s2]

Figure

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Answer: (b)

  1. A moves with B only through friction: amax=μ1g=0.2×10=2a_{max} = \mu_1 g = 0.2 \times 10 = 2 m/s2^2.
  2. Table friction on B: μ2(mA+mB)g=0.2×4×10=8\mu_2(m_A + m_B)g = 0.2 \times 4 \times 10 = 8 N
  3. Whole system: F−8=(mA+mB)a=4×2F - 8 = (m_A + m_B)a = 4 \times 2
  4. F=16F = 16 N, so the answer is (b).
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