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Answer to Question #3087 in Mechanics | Relativity for TAHIRA

Question #3087
A BOOK OF WEIGHT 2.5N RESTS ON A HORIZONTAL TABLE TOP AND BY MEANS OF A PENCIL A HORIZONTAL FORCE OF 2N IS APPLIED TO THE BOOK CAUSING IT TO SLIDE.THE FORCE OF FRICTION ON THE SLIDING BOOK IS 1N.
A-DRAW A DIAGRAM OF THE SET UP TO SHOW THE BOOK AT REST ON THE TABLE TOP AND FORCES ACTED ON THE BOOK BEFORE THE PENCIL WAS USED.
B-NAME AND GIVE THE MAGNITUDE OF EACH FORCE ACTING ON THE BOOK BEFORE THE FORCE WAS APPLIED WITH THE PENCIL.
C-WRITE A BRIEF EXPLANATION FOR THE STAE OF REST OF THE BOOK.
D-DRAW A SEPARATE DIAGRAM TO SHOW THE SLIDING BOOK AND THE FORCES ACTED ON IT.
E-CALCULATE THE VELOCITY OF THE BOOK WHEN THE APPLIED FORCE HAS BEEN ACTING ON IT FOR 1.5 SECONDS.
USE GRAVITY=10 (N/KG)
Expert's answer
Vector sum of horizonal forces is
Fpencil - Ffriction = 2[N] - 1[N] = 1 [N]
The mass of the book is 2.5[N] /10 [N/kg] = 0.25 kg
The acceleration is a = F/m = 1[N]/0.25[kg] = 4 [N/kg]
After 1.5 seconds the bool will have V = at2/2 = 4* 1.52 /2 = 4.5 [m/s].

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