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: 46

Answer

Please see the work below.

Work Step by Step

We know that $a=\frac{2y}{t^2}$ We plug in the known values to obtain: $a=\frac{2(10.8)}{(3.00)^2}=2.40\frac{m}{s^2}$ Now we can find the force $F=ma=2.50(2.40)=6.00N$ As $F=F_1+F_2$ $\implies F_2=F-F_1$ We plug in the known values to obtain: $F_2=6.00-15=-9.00jN$
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