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21 September, 13:24

84. Two blocks, one of mass 5.0 kg and the other of

mass 3.0 kg, are tied together with a massless rope

as in Figure 28. This rope is strung over a massless,

resistance-free pulley. The blocks are released from

rest. Find the following:

a. the tension in the rope

b. the acceleration of the blocks

Hint: You will need to solve two simultaneous equations.

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Answers (1)
  1. 21 September, 13:56
    0
    a) 37 N

    b) 2.5 m/s²

    Explanation:

    Let's say m₁ is the 5 kg mass, and m₂ is the 3 kg mass.

    Draw a free body diagram for each block. For both, there are two forces: weight down and tension up.

    Apply Newton's second law to the 5 kg mass in the y direction (remember it is accelerating downward).

    ∑F = ma

    T - m₁g = m₁ (-a)

    m₁g - T = m₁a

    Apply Newton's second law to the 3 kg mass in the y direction:

    ∑F = ma

    T - m₂g = m₂a

    a) Solve the system of equations. First, solve for a in each equation, then set equal and solve for T.

    a = (m₁g - T) / m₁

    a = (T - m₂g) / m₂

    (m₁g - T) / m₁ = (T - m₂g) / m₂

    m₂ (m₁g - T) = m₁ (T - m₂g)

    m₁m₂g - Tm₂ = m₁T - m₁m₂g

    2m₁m₂g = (m₁ + m₂) T

    T = 2m₁m₂g / (m₁ + m₂)

    Given m₁ = 5 kg and m₂ = 3 kg:

    T = 2 (5 kg) (3 kg) (9.8 m/s²) / (5 kg + 3 kg)

    T = 36.75 N

    Rounded to two significant figures, the tension is 37 N.

    b) Solve the system of equations. First, solve for T in each equation, then set equal and solve for a.

    T = m₁g - m₁a

    T = m₂g + m₂a

    m₁g - m₁a = m₂g + m₂a

    m₁g - m₂g = m₁a + m₂a

    (m₁ - m₂) g = (m₁ + m₂) a

    a = (m₁ - m₂) g / (m₁ + m₂)

    Given m₁ = 5 kg and m₂ = 3 kg:

    a = (5 kg - 3 kg) (9.8 m/s²) / (5 kg + 3 kg)

    a = 2.45 m/s²

    Rounded to two significant figures, the magnitude of the acceleration of the blocks is 2.5 m/s².
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