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

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 + at²

we substitute;

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

1.29 = 0×0.850  + ×a×(0.850)²

1.29 = 0.36125a

a = 1.29 / 0.36125

a = 3.5709 m/s²

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

T₁ = m₁a

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

a 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

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

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