Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 7

Answer

(a) $F_{elec} = 0.0036~N$ (b) $T = 0.0104~N$

Work Step by Step

The vertical component of the tension $T_y$ is equal in magnitude to the ball's weight. We can find the horizontal component of the tension $T_x$. So; $\frac{T_x}{T_y} = tan(\theta)$ $T_x = T_y~tan(\theta)$ $T_x = mg~tan(\theta)$ $T_x = (0.0010~kg)(9.80~m/s^2)~tan(20^{\circ})$ $T_x = 0.0036~N$ The force $F_{elec}$ is equal in magnitude to the horizontal component of the tension $T_x$. So; $F_{elec} = T_x$ $F_{elec} = 0.0036~N$ (b) We can find the tension $T$ in the string. $cos(\theta) = \frac{T_y}{T}$ $T = \frac{T_y}{cos(\theta)}$ $T = \frac{mg}{cos(\theta)}$ $T = \frac{(0.0010~kg)(9.80~m/s^2)}{cos(20^{\circ})}$ $T = 0.0104~N$
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