Answer:
The coefficient of friction between m₂ and the table is approximately 0.2703
The magnitude of the tension in cable is 28.737 N
Explanation:
The given parameters are;
The mass of the block on the table, m₂ = 5.8 kg
The mass of the hanging block, m₁ = 4.5 kg
The distance in which the block has moved = 0.317 m
The speed after the block moves = 1.33 m/s
The acceleration due to gravity, g = 9.8 m/s²
The coefficient of friction between m₂ and the table = μ
The acceleration, a, of the blocks is given from the formula of speed, v, of a body at a given distance, s, from rest as follows;
v² = 2 × a × s
Where;
v = 1.33 m/s
s = 0.317 m
Substituting the values gives;
1.33² = 2 × a × 0.317
a = 1.33²/(2 × 0.317) ≈ 2.79
The acceleration, a ≈ 2.79 m/s²
Therefore, we have the force acting on the system given as follows;
Weight of m₁ - Frictional force of m₂ = (m₁ + m₂) × a
4.5 × 9.8 - 5.8 × 9.8 × μ = (4.5 + 5.8) × 2.79
4.5 × 9.8 - 5.8 × 9.8 × μ = 28.737 N
(4.5 × 9.8)N - 28.737 N = 5.8 × 9.8 × μ
44.1 N - 28.737 N =56.84 N × μ
15.363 N = 56.84 N × μ
μ = 15.363 N/(56.84 N) ≈ 0.2703
The coefficient of friction between m₂ and the table = μ ≈ 0.2703
The coefficient of friction between m₂ and the table ≈ 0.2703
The magnitude of the tension in cable = (m₁ + m₂) × a = 28.737 N.
The magnitude of the tension in cable = 28.737 N
