Question

A textbook of mass 2.09 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.190 m, to a hanging book with mass 3.08 kg. The system is released from rest, and the books are observed to move a distance 1.26 m over a time interval of 0.800 s.

Answer 1

The tension in the part of the cord attached to the textbook is 6.58 N.

Step-by-step 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

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

`a=(2y)/(t^2)`

Put the value into the formula

`a=(2*1.26)/(0.800)`

`a=3.15\ m/s^2`

We need to calculate the tension

When the book is moving with acceleration

Using formula of tension

`T=ma`

`T=2.09*3.15`

`T=6.58\ N`

Hence, The tension in the part of the cord attached to the textbook is 6.58 N.