Two blocks are suspended from opposite ends of a light rope that passes over a light, frictionless pulley. One block has mass m1 and the other has mass m2, where m2>m1. The two blocks are released from rest, and the block with mass m2 moves downward 6.00 m in 2.00 s after being released. While the blocks are moving, the tension in the rope is 18.0 N.

Part A

Calculate m1.

Express your answer with the appropriate units.

Part B

Calculate m2.

Express your answer with the appropriate units.

a) m₁ = 1.41 kg, b) m₂ = 2.65 kg

Explanation:

For this exercise we will use Newton's second law

Block 1

T - W₁ = m₁ a

Block 2

W₂ - T = m₂ a

We have selected the positive block 1 rising and block two lowering, as the pulley has no friction does not affect the movement

Let's use kinematics to look for acceleration

y = v₀ t + ½ a t²

As part of the rest the initial speed is zero

a = 2 y / t²

a = 2 6.00 / 2²

a = 3 m / s²

Let's replace in the equation of block 1

a) T = m₁ g + m₁ a

m₁ = T / (g + a)

m₁ = 18.0 / (9.8 + 3)

m₁ = 1.41 kg

b) we substitute in the equation of block 2

W₂ - T = m₂ a

m₂ g - m₂ a = T

m₂ = T / (g-a)

m₂ = 18.0 / (9.8 - 3)

m₂ = 2.65 kg