#### Answer

(a) The book slides a distance of 0.67 meters.
(b) The book will slide back down the slope.

#### Work Step by Step

We can find the acceleration of the system of the cup and the book. Let $m_c$ be the mass of the cup. Let $m_b$ be the mass of the book.
$\sum F = (m_b+m_c)~a$
$m_c~g+m_b~g~sin(\theta)+m_b~g~cos(\theta)~\mu_k = (m_b+m_c)~a$
$a = \frac{m_c~g+m_b~g~sin(\theta)+m_b~g~cos(\theta)~\mu_k}{m_b+m_c}$
$a = \frac{(0.500~kg)(9.80~m/s^2)+(1.0~kg)(9.80~m/s^2)~sin(20^{\circ})+(1.0~kg)(9.80~m/s^2)~cos(20^{\circ})(0.20)}{1.0~kg+0.500~kg}$
$a = 6.73~m/s^2$
The acceleration of the book is $6.73~m/s^2$ directed down the slope.
We can find the distance $d$ that the book slides.
$d = \frac{v^2-v_0^2}{2a}$
$d = \frac{0-(3.0~m/s)^2}{(2)(-6.73~m/s^2)}$
$d = 0.67~m$
The book slides a distance of 0.67 meters.
(b) When the book comes to a stop, we can find the maximum possible force of static friction (directed up the slope) which can be exerted on the book.
$F_f = m_b~g~cos(\theta)~\mu_s$
$F_f = (1.0~kg)(9.80~m/s^2)~cos(20^{\circ})(0.50)$
$F_f = 4.6~N$
We can find the force on the book directed down the slope if the book is at rest. Note that when the system is at rest, the tension in the string is equal to the cup's weight.
$F_{down} = m_b~g~sin(\theta)+T$
$F_{down} = m_b~g~sin(\theta)+m_c~g$
$F_{down} = (1.0~kg)(9.80~m/s^2)~sin(20^{\circ})+(0.500~kg)(9.80~m/s^2)$
$F_{down} = 8.3~N$
Since the force on the book which is directed down the slope is greater in magnitude than the maximum possible force of static friction directed up the slope, the book will slide back down the slope.