Rankers Physics
Topic: Laws of Motion
Subtopic: Friction

The minimum value of coefficient of friction (μ) such that block of mass ‘5 kg’ remains at rest is
Image related to
0.3
0.4
0.5
0.6

Solution:

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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