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A textbook of mass m1 = 1.92 kg rests on a frictionless, horizontal surface. A cord attached to the book runs horizontally to a pulley whose diameter is 0.200 m , then down to a hanging book with mass m2 = 2.93 kg . The system is released from rest, and the books are observed to move a distance 1.17 m over a time interval of 0.810 s. What is the tension in the part of the cord attached to the textbook?

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Answer

given,

mass of text book = m₁ = 1.92 Kg

diameter of pulley = 0.2 m

mass of the hanging book = m₂ = 2.93 Kg

book is observed to move = 1.17 m

time interval = 0.810 s

tension in the cord attached = ?

computing the horizontal forces

∑ Fx = 0

T₁ = m₁ a₁

using equation of motion

[tex]s = ut + \dfrac{1}{2}at^2[/tex]

[tex]a= \sqrt{2 \times 1.17 \times 0.81^2}[/tex]

a = 1.24 m/s²

now,

T₁ = 1.92 x 1.24

T₁ = 2.38 N

now along y - axis

[tex]m_2g - T_2 = m_2 a_2[/tex]

[tex]a_2 = a_1[/tex]

[tex]T_2 = m_2 (g -a_2)[/tex]

[tex]T_2 = 2.93(9.8 - 1.24)[/tex]

T₂  = 25.08 N

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