Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 4 - Exercises and Problems: 57

Answer

Please see the work below.

Work Step by Step

We know that $F_2=Kx$ We plug in the known values to obtain: $F_2=(8100)(0.051)=413.1N$ Now $a_2=\frac{F_2}{m_2}$ $a_2=a=\frac{413.1}{4900}=0.84306\frac{m}{s^2}$ Hence, the net force is given as $F=(m_1+m_2)a$ We plug in the known values to obtain: $F=(640+490)(0.84306)=950N$
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