University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 5 - Applying Newton's Laws - Problems - Exercises - Page 165: 5.68


The tension in the rope is 29.7 N.

Work Step by Step

$v(t) = (2.00~m/s^2)~t+(0.600~m/s^3)~t^2$ $a(t) = (2.00~m/s^2)+(1.200~m/s^3)~t$ We can find the time $t$ when $v = 9.00~m/s$ $v = (2.00~m/s^2)~t+(0.600~m/s^3)~t^2 = 9.00~m/s$ $(0.600~m/s^3)~t^2 + (2.00~m/s^2)~t- 9.00~m/s = 0$ We can use the quadratic formula to find $t$. $t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{-(2.00)\pm \sqrt{(2.00)^2-(4)(0.600)(-9.00)}}{(2)(0.600)}$ $t = -5.88~s, 2.55~s$ Since the negative value is unphysical, the solution is $t = 2.55~s$. When t = 2.55 s: $a = (2.00~m/s^2)+(1.200~m/s^3)(2.55~s)$ $a = 5.06~m/s^2$ We can use a force equation to find the tension $T$ in the rope. $\sum F = ma$ $T - mg = ma$ $T = ma + mg = (2.00~kg)(5.06~m/s^2)+(2.00~kg)(9.80~m/s^2)$ $T = 29.7~N$ The tension in the rope is 29.7 N.
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