## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 162: 29

#### 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 $T_x$. $\frac{T_x}{T_y} = tan(\theta)$ $T_x = T_y~tan(\theta)$ $T_x = mg~tan(\theta)$ $T_x = (0.0010~kg)(9.80~m/s^2)~tan(20^{\circ})$ $T_x = 0.0036~N$ The force $F_{elec}$ is equal in magnitude to the horizontal component of the tension $T_x$. $F_{elec} = T_x$ $F_{elec} = 0.0036~N$ (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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