Answer:
Approximately [tex]11.0\; \rm m \cdot s^{-1}[/tex]. (Assuming that [tex]g = 9.81 \; \rm N \cdot kg^{-1}[/tex], and that the tabletop is level.)
Explanation:
Weight of the book:
[tex]W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N[/tex].
If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, [tex]F(\text{normal force}) \approx 10.202\; \rm N[/tex].
As a side note, the [tex]F_N[/tex] and [tex]W[/tex] on this book are not equal- these two forces are equal in size but point in the opposite directions.
When the book is moving, the friction [tex]F(\text{kinetic friction})[/tex] on it will be equal to
That is:
[tex]\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}[/tex].
Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that [tex]15.0\; \rm N[/tex] applied force. The net force on the book shall be:
[tex]\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}[/tex].
Apply Newton's Second Law to find the acceleration of this book:
[tex]\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}[/tex].
