Answer
(a) $T_x = 7.50~N$
$T_y = 18.2~N$
(b) $I = 0.016~kg~m^2$
Work Step by Step
(a) We can find the acceleration of the system.
$d = \frac{1}{2}at^2$
$a = \frac{2d}{t^2}$
$a = \frac{(2)(1.20~m)}{(0.800~s)^2}$
$a = 3.75~m/s^2$
Let $T_y$ be the tension in the part of the string that is vertical. Let $m_1$ be the mass of the book that is hanging.
$\sum F = m_1a$
$m_1g-T_y = m_1a$
$T_y = m_1(g-a)$
$T_y = (3.00~kg)(9.80~m/s^2-3.75~m/s^2)$
$T_y = 18.2~N$
Let $T_x$ be the tension in the part of the string that is horizontal. Let $m_2$ be the mass of the book that is on the surface.
$T_x = m_2a$
$T_x = (2.00~kg)(3.75~m/s^2)$
$T_x = 7.50~N$
(b) We can find the moment of inertia of the pulley.
$\sum \tau = I~\alpha$
$T_y~R-T_x~R = I \frac{a}{R}$
$I = \frac{R^2(T_y-T_x)}{a}$
$I = \frac{(0.075~m)^2~(18.2~N-7.50~N)}{3.75~m/s^2}$
$I = 0.016~kg~m^2$
