Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - Questions - Page 131: 19

Answer

(a) The normal force exerted by the road on the car is 5970 N. (b) The normal force exerted by the car on the driver is 379 N. (c) The normal force on the driver is zero when the speed is 29.4 m/s.

Work Step by Step

(a) $\sum F = m \frac{v^2}{r}$ $F_g - F_N = m \frac{v^2}{r}$ $F_N = F_g - m \frac{v^2}{r}$ $F_N = (975 ~kg)(9.80 ~m/s^2) - (975 ~kg)\frac{(18.0 ~m/s)^2}{(88.0 ~m)}$ $F_N = 5970 ~N$ The normal force exerted by the road on the car is 5970 N. (b) (a) $\sum F = m \frac{v^2}{r}$ $F_g - F_N = m \frac{v^2}{r}$ $F_N = F_g - m \frac{v^2}{r}$ $F_N = (62.0 ~kg)(9.80 ~m/s^2) - (62.0 ~kg)\frac{(18.0 ~m/s)^2}{(88.0 ~m)}$ $F_N = 379 ~N$ The normal force exerted by the car on the driver is 379 N. (c) If the normal force on the driver is 0, then $F_g$ must be exactly equal to the required centripetal force. $m \frac{v^2}{r} = mg$ $v = \sqrt{gr}$ $v = \sqrt{(9.80 ~m/s^2)(88.0 ~m)}$ $v = 29.4 ~m/s$ The normal force on the driver is zero when the speed is 29.4 m/s.
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