Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 203: 14a


2.98 m/s

Work Step by Step

Since only conservative forces are acting on the ball, the mechanical energy $E_{mec}$ is conserved and so $ΔE_{mec} = ΔK + ΔU= \frac{1}{2} m(v^2–v_{0}^{2})+mgΔh=0$. The problem is asking for initial speed, so solving for $v_{0}$ gives: $ \frac{1}{2}m(v^{2} – v_{0}^{2}) + mg\Delta h = 0$ $\frac{1}{2}m(v^{2} – v_{0}^{2}) = – mg\Delta h$ $v^{2} – v_{0}^{2} = – 2g\Delta h$ $ v_{0}^{2} = v^{2} + 2g\Delta h$ $v_{0} = \sqrt {v^{2} + 2g\Delta h}$ The ball is to reach the vertically upward position with zero speed, so $v$= 0. Looking at the picture we see that the change in height is the same as the rod’s length, and so $Δh$ = L = 0.452 m. The problem also gives $m$ = 0.341 kg and we know that $g$ = 9.8 m/s². So plugging in our values, we get: $v_{0} = \sqrt {v^{2} + 2g\Delta h}$ = $\sqrt {0^{2} + 2(9.8 m/s²)(0.452 m)}$ $\approx$ 2.98 m/s
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