University Physics with Modern Physics (14th Edition)

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

Chapter 23 - Electric Potential - Problems - Exercises - Page 778: 23.13

Answer

$v_B = 7.41 \,\text{m/s}$

Work Step by Step

We will use the conservation energy law to get the speed \begin{gather} K_{A}+U_{A}= K_{B}+U_{B} \\ \frac{1}{2} m v_{A}^{2}+ qV_{A} =\frac{1}{2} m v_{B}^{2}+qV_{B} \\ \frac{1}{2} m v_{B}^{2} =\frac{1}{2} m v_{A}^{2}+q( V_{A}-V_{B})\\ \frac{1}{2} m v_{B}^{2} = \frac{1}{2} (2.00 \times 10^{-4} \mathrm{kg}) (5.00 \mathrm{~m/s})^{2}+5 \times 10^{-6} \,\text{C}( +200 \mathrm{V} -800 \mathrm{V})\\ \frac{1}{2} m v_{B}^{2} = 0.0055 \,\text{J} \end{gather} The speed will be \begin{align} v_B &= \sqrt{\dfrac{2 (0.0055 \,\text{J})}{m}}\\ &= \sqrt{\dfrac{2 (0.0055 \,\text{J})}{2.00 \times 10^{-4} \mathrm{kg}}}\\ &= \boxed{7.41 \,\text{m/s}} \end{align}
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