Consider the situation shown in figure (8-E2). The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 m/s after it has descended through a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.
Mass of the block on the table, m1 = 4 kg
Mass of the block hanging from the table, m2 = 1 kg
Acceleration of the system = a
Initially when the system was at rest, speed u = 0 m/s
Speed at a later time, v = 0.3 m/s
Distance travelled, s = 1 m
From equation of motion, 2as = v2-u2
⇒ 2×a×1 = 0.3×0.3 - 0
⇒ a = 0.045 m/s2
For the 1 kg block hanging from the table, Let the tension in the string be 2T
m2a = m2g – 2T
2T = 1×(0.045 + 9.8)
2T = 9.845
T = 4.9225 N
For the 4kg block on the table, Let the force of friction be F and tension in the string will be T
T – F = 2m1a
If coefficient of friction is μ, F = μm1g
T – μm1g = 2m1a
4×μ×9.8 = 4.9225 - 2×4×0.045
39.2μ = 4.5625
μ = 0.116 ≈ 0.12
∴ Coefficient of friction between the block and the table = 0.12