Question

A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100 m , to a hanging book with mass 3.10 kg . The system is released from rest, and the books are observed to move a distance 1.29 m over a time interval of 0.850 s . Part A What is the tension in the part of the cord attached to the textbook

Answer 1

the tension in the part of the cord attached to the textbook is 7.4989 N

Explanation:

Given the data in the question;

As illustrated in the image below;

first we determine the value of the acceleration,

along vertical direction; we use the second equation of motion;

y = ut +
`(1)/(2)`a
`_y`t²

we substitute;

0 m/s for u, 1.29 m for y, 0.850 s for t,

1.29 = 0×0.850 +
`(1)/(2)`×a
`_y`×(0.850)²

1.29 = 0.36125a
`_y`

a
`_y` = 1.29 / 0.36125

a
`_y` = 3.5709 m/s²

Now when the text book is moving with acceleration , the dynamic equation will be;

T₁ = m₁a
`_y`

where m₁ is the mass of the text book ( 2.10 kg )

a
`_y` is the vertical acceleration ( 3.5709 m/s² )

so we substitute

T₁ = 2.10 × 3.5709

T₁ = 7.4989 N

Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N