A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point B it has slowed to 1.25 m/s. (a) How much work was done on the book between A and B? (b) If - 0.750 J of work is done on the book from B to C, how fast is it moving at point C? (c) How fast would it be moving at C if + 0.750 J of work was done on it from B to C?

Solution 14E Step 1: We need to find work done between point A and B It is given by Using K = 1/2 mv 2 Since we have two points we shall find the difference in kinetic energy between A and B We have v = 3.21 m/s A vB= 1.25 m/s m = 1.50 kg Kinetic energy at A K = 1/2 mv 2 A A 2 KA= (1/2) × 1.50 kg × (3.21 m/s) KA= 7.72 J Similarly Kinetic energy at B K = 1/2 mv 2 B B 2 KB= (1/2) × 1.50 kg × (1.25 m/s) KB= 1.17 J We get work done as W = K K B A W = 1.17 J 7.72 J W = 6.55 J Hence we have work done as 6.55 J Step 2 : We need to find the velocity of the book at point C It is given by Kc= K B W It is given work done is W = 750 J We have K = 1.17 J B We can resolve to find V = (1/2 mv ) W c B Substituting we get V c (1.25 ) + ( 0.750 × 2)/1.50 kg V = 0.75 m/s c Hence we get velocity at point C as 0.75 m/s