Two blocks are arranged at the ends of a massless cord over a frictionless massless pulley as shown in the figure. Assume the system starts
from rest. When the masses have moved a distance of 0.317 m, their speed is 1.33 m/s.
The acceleration of gravity is 9.8 m/s
2

What is the coefficient of friction between
m2 and the table?

What is the magnitude of the tension in the
cord?
Answer in units of N.

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

explanation:

newton's third law, for every action there is an equal and opposite reaction.