I assume friction is the only force acting on the book as it slides.
(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:
<em>W</em> = ∆<em>K</em>
<em>W</em> = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²
<em>W</em> ≈ -6.56 J
(B) Using the work-energy theorem again, the speed <em>v</em> of the book at point C is such that
-0.750 J = 1/2 (1.50 kg) <em>v</em> ² - 1/2 (1.50 kg) (1.25 m/s)²
==> <em>v</em> = 0.750 m/s
(C) Take the left side to be positive, then solve again for <em>v</em>.
0.750 J = 1/2 (1.50 kg) <em>v</em> ² - 1/2 (1.50 kg) (1.25 m/s)²
==> <em>v</em> ≈ 1.60 m/s