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16 January, 00:44

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.

+2
Answers (1)
  1. 16 January, 01:00
    0
    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
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