Answer:
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 + [tex]\frac{1}{2}[/tex]a[tex]_y[/tex]t²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + [tex]\frac{1}{2}[/tex]×a[tex]_y[/tex]×(0.850)²
1.29 = 0.36125a[tex]_y[/tex]
a[tex]_y[/tex] = 1.29 / 0.36125
a[tex]_y[/tex] = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a[tex]_y[/tex]
where m₁ is the mass of the text book ( 2.10 kg )
a[tex]_y[/tex] 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
