Question

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?

Answer 1

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

`s = ut + (1)/(2)at^2`

`a= √(2 * 1.17 * 0.81^2)`

a = 1.24 m/s²

now,

T₁ = 1.92 x 1.24

T₁ = 2.38 N

now along y - axis

`m_2g - T_2 = m_2 a_2`

`a_2 = a_1`

`T_2 = m_2 (g -a_2)`

`T_2 = 2.93(9.8 - 1.24)`

T₂ = 25.08 N