The minimum value of coefficient of friction (μ) such that block of mass ‘5 kg’ remains at rest is 0.30.40.50.6Solution:Given m1=5m_1 = 5 kg, m2=3m_2 = 3 kg, and g=10g = 10 m/s²: Tension: T=m2g=3×10=30T = m_2 g = 3 \times 10 = 30 N Friction force: f=μN=μ×50f = \mu N = \mu \times 50 At equilibrium: μ×50=30\mu \times 50 = 30 Solving, μ=3050=0.6\mu = \frac{30}{50} = 0.6 .
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