Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 181: 42

Answer

(a) stay the same (b) by a factor of 2.

Work Step by Step

(a) We know that the acceleration of the system is given as $a=(\frac{m_2}{m_1+m_2})g$ But given that $m_1=2m_1$and $m_2=2m_2$ $\implies a_{new}=(\frac{2m_2}{2m_1+2m_2})g$ $\implies a_{new}=\frac{2}{2}(\frac{m_2}{m_1+m_2})g$ $\implies a_{new}=a$ Thus, the acceleration will stay the same. (b) We know that the tension is given as $T=(\frac{m_1m_2}{m_1+m_2})g$ But $m_1=2m_1$ and $m_2=2m_2$ $\implies T_{new}=(\frac{(2m_1)(2m_2)}{2m_1+2m_2})g$ $\implies T_{new}=\frac{4}{2}(\frac{m_1m_2}{m_1+m_2})g$ $\implies T_{new}=2T$ Thus, the tension in the string is increased by a factor of 2.
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