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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.

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1 vote

Answer:

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.

User Mweber
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