Answer:
The tension in the part of the cord attached to the textbook is 6.58 N.
Explanation:
Given that,
Mass of textbook = 2.09 kg
Diameter = 0.190 m
Mass of hanging book = 3.08 kg
Distance = 1.26 m
Time interval = 0.800 s
Suppose,we need to calculate the tension in the part of the cord attached to the textbook
We need to calculate the acceleration
Using equation of motion
[tex]y=ut+\dfrac{1}{2}at^2[/tex]
[tex]a=\dfrac{2y}{t^2}[/tex]
Put the value into the formula
[tex]a=\dfrac{2\times1.26}{0.800}[/tex]
[tex]a=3.15\ m/s^2[/tex]
We need to calculate the tension
When the book is moving with acceleration
Using formula of tension
[tex]T=ma[/tex]
[tex]T=2.09\times3.15[/tex]
[tex]T=6.58\ N[/tex]
Hence, The tension in the part of the cord attached to the textbook is 6.58 N.
