Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 7 - Newton's Third Law - Exercises and Problems - Page 189: 52

Answer

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Work Step by Step

a) From the given figures, it seems that the two forces of 9.8 N (exerted on the second block of mass 2 kg) and 2.94 N (exerted on the first block of mass 1 kg) are friction forces. This means that the two objects are actually moving toward the right while the friction forces oppose them. We can imagine two boxes stuck together and pushed to the right on a horizontal surface and then we leave them to slide without any pushing forces. Now the two blocks are sliding to the right at some acceleration of $a_x$ what is its magnitude? And find the action-reaction force $F$. b) Finding the acceleration of the system. $$\sum F_{x1}=-f_{k1}-F=-m_1a_x$$ Thus, $$m_1a_x=f_{k1}+F\tag 1$$ $$\sum F_{x2}=F-f_{k2} =-m_2a_x$$ Thus, $$F=f_{k2} -m_2a_x\tag 2$$ Plugging into (1); $$m_1a_x=f_{k1}+f_{k2} -m_2a_x $$ Solving for $a_x$; $$m_1a_x+m_2a_x=f_{k1}+f_{k2} $$ $$a_x =\dfrac{f_{k1}+f_{k2} }{m_1 +m_2 }$$ Plugging the known; $$a_x =\dfrac{ 2.94+9.8 }{1+2}=\color{red}{\bf 4.25 }\;\rm m/s^2$$ Plugging into (2) to find $F$; $$F=f_{k2} -4.25 m_2 $$ Plugging the known; $$F=9.8 -(4.25 \times2)= \color{red}{\bf1.3}\;\rm N$$
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