stacked pair of books with masses m1= 2.0 kg (bottom book) and m2 = 1.5 kg (top book) are tossed onto a table. The books strike the table with no vertical velocity and their common horizontal speed is ?o = 0.75 m/s. The kinetic friction coefficient between the bottom book and the table is ?k1=0.45; the kinetic and static friction coefficients between the two books are ?k2=0.3 and ?s2= 0.4. Find the final horizontal position of each book relative to the spot where the stack hits the table.
stacked pair of books with masses m1= 2.0 kg (bottom book) and m2 = 1.5 kg (top book) are tossed onto a table. The books strike the table with no vertical velocity and their common horizontal speed is ?o = 0.75 m/s. The kinetic friction coefficient between the bottom book and the table is ?k1=0.45; the kinetic and static friction coefficients between the two books are ?k2=0.3 and ?s2= 0.4. Find the final horizontal position of each book relative to the spot where the stack hits the table.
Glencoe Physics: Principles and Problems, Student Edition
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ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter5: Displacement And Force In Two Dimensions
Section: Chapter Questions
Problem 8STP
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Question
A stacked pair of books with masses m1= 2.0 kg (bottom book) and m2 = 1.5 kg (top book) are tossed onto a table. The books strike the table with no vertical velocity and their common horizontal speed is ?o = 0.75 m/s.
The kinetic friction coefficient between the bottom book and the table is ?k1=0.45; the kinetic and static
friction coefficients between the two books are ?k2=0.3 and ?s2= 0.4. Find the final horizontal
position of each book relative to the spot where the stack hits the table.
Expert Solution
Step 1
The normal force exerted by the table on the first bolook N1= its weight = mg= 2×10=20 N .
Similarly N2= 1.5×10=15 N .
So frictional forces f1=k1N1= 0.45 ×20 = 9 N .
So f2= k2N2= 0.3×15 = 4.5 N .
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