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

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 41

Answer

The tangential acceleration of car A is $9.80~m/s^2$ The tangential acceleration of car B is $12.9~m/s^2$ The tangential acceleration of car C is $6.68~m/s^2$

Work Step by Step

Note that the force of kinetic friction causes each car to decelerate. Let $M$ be the mass of each car. We can find the tangential acceleration of car A; $F_f = Ma$ $F_N~\mu_k = Ma$ $Mg~\mu_k = Ma$ $a = g~\mu_k$ $a = (9.80~m/s^2)(1.0)$ $a = 9.80~m/s^2$ The tangential acceleration of car A is $9.80~m/s^2$. We can use the centripetal force to find the normal force $F_N$ of the road on car B; $\sum F = \frac{Mv^2}{r}$ $F_N-Mg = \frac{Mv^2}{r}$ $F_N = \frac{Mv^2}{r}+Mg$ $F_N = M~(\frac{v^2}{r}+g)$ $F_N = M~(\frac{(25~m/s)^2}{200~m}+9.80~m/s^2)$ $F_N = M~(12.9~m/s^2)$ We can find the tangential acceleration of car B. $F_f = Ma$ $F_N~\mu_k = Ma$ $M~(12.9~m/s^2)~\mu_k = Ma$ $a = (12.9~m/s^2)~\mu_k$ $a = (12.9~m/s^2)(1.0)$ $a = 12.9~m/s^2$ The tangential acceleration of car B is $12.9~m/s^2$. We can use the centripetal force to find the normal force $F_N$ of the road on car C; $\sum F = \frac{Mv^2}{r}$ $Mg-F_N = \frac{Mv^2}{r}$ $F_N = Mg - \frac{Mv^2}{r}$ $F_N = M~(g-\frac{v^2}{r})$ $F_N = M~(9.80~m/s^2 - \frac{(25~m/s)^2}{200~m})$ $F_N = M~(6.68~m/s^2)$ We can find the tangential acceleration of car C. $F_f = Ma$ $F_N~\mu_k = Ma$ $M~(6.68~m/s^2)~\mu_k = Ma$ $a = (6.68~m/s^2)~\mu_k$ $a = (6.68~m/s^2)(1.0)$ $a = 6.68~m/s^2$ The tangential acceleration of car C is $6.68~m/s^2$.
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