## University Physics with Modern Physics (14th Edition)

(a) $T_x = 7.50~N$ $T_y = 18.2~N$ (b) $I = 0.016~kg~m^2$
(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. $\sum \tau = I~\alpha$ $T_y~R-T_x~R = I \frac{a}{R}$ $I = \frac{R^2(T_y-T_x)}{a}$ $I = \frac{(0.075~m)^2~(18.2~N-7.50~N)}{3.75~m/s^2}$ $I = 0.016~kg~m^2$