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 11 - Work - Exercises and Problems - Page 307: 67

Answer

a) $6.53\;\rm m/s^2$ b) $11.7\;\rm m/s$ c) $1.79\;\rm s$

Work Step by Step

a) We know that the total weight exerted on the drive wheels is $\frac{2}{3}m_{car}g$. This is also the same normal force exerted on them since the car has no motion in the vertical direction. $$\sum F_y=F_n-m_{wheels}g=m_{wheels}a_y=m_{wheels}(0)=0$$ Thus, $$F_n=m_{wheels}g=\frac{2}{3}m_{car}g\tag 1$$ We know that the only force exerted on the drive wheels horizontally is the static friction force between the tires and the road. This is the only force that pushes the car forward. $$\sum F_x=f_s=m_{car}a_x$$ Thus, $$\mu_sF_n=m_{car}a_x $$ Plugging from (1); $$ \frac{2}{3}\mu_sm_{car}g=m_{car}a_x$$ Thus, $$ \frac{2}{3} \mu_s\color{red}{\bf\not} m_{car}g=\color{red}{\bf\not} m_{car}a_x$$ $$a_x= \frac{2}{3}\mu_sg$$ Plugging the known; $$a_x= \frac{2}{3}\times 1\times 9.8=\color{red}{\bf 6.53}\;\rm m/s^2$$ This is the maximum acceleration of the car. ________________________________________________ b) We know that the power output is given by $$P=Fv$$ Thus, the maximum speed is given $$v_{max}=\dfrac{P}{F}$$ Recall that the force that pushed the car forward is the static friction force. Thus, $$v_{max}=\dfrac{P}{f_s}=\dfrac{0.7P_{tot}}{ \frac{2}{3} \mu_sm_{car}g}$$ Plugging the known; $$v_{max}= \dfrac{0.7\times (217\times 746)}{ \frac{2}{3} \times 1\times 1480\times 9.8 }=\color{red}{\bf 11.7}\;\rm m/s$$ ________________________________________________ c) Reaching the maximum power output means reaching the maximum speed. Thus, we can use the kinematic formula of $$v_{max}=v_i+a_xt$$ The car starts from rest, so $v_i=0$. Solving for $t$; $$t=\dfrac{v_{max}}{a_x}=\dfrac{11.7}{6.53}$$ $$t=\color{red}{\bf 1.79}\;\rm s$$
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