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 10 - Energy - Exercises and Problems - Page 275: 61

Answer

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Work Step by Step

a) We will work with the assumption the author set. And we can assume that the glider's energy is conserved. Thus, $$E_i=E_f$$ $$K_1+U_i=K_f+U_f$$ we know that the glider starts from rest which means that its initial kinetic energy is zero. $$0+U_i=K_f+U_f$$ Plugging $U=\dfrac{c}{x}$ $$\dfrac{c}{x_i}=\frac{1}{2}mv^2+\dfrac{c}{x_f}$$ Solving for $v^2$; $$\dfrac{c}{x_i}-\dfrac{c}{x_f}=\frac{1}{2}mv^2$$ $$v^2= \dfrac{\dfrac{2c}{x_i}-\dfrac{2c}{x_f}}{m} $$ $$v^2= \dfrac{2c}{x_i m}-\dfrac{2c}{x_fm} $$ $$v^2= \dfrac{2c}{x_i m}-\dfrac{2c}{x_fm} $$ The final position is at $x=1$, so $$\boxed{v^2= \dfrac{2c}{m}x_i^{-1}-\dfrac{2c}{ m} } $$ This equation is similar to the straight line equation $y=mx+b$ where $m$ is the slope of the straight line, and $b$ is the interception point with $y$-axis. Now we need to draw $v^2$ versus $\dfrac{1}{x_i}$, by plugging the data in the table below. \begin{array}{|c|c| } \hline \dfrac{1}{x}\;{\rm (m^{-1})}& v^2\;{\rm (m^2/s^2)} \\ \hline \dfrac{1}{0.02}=50& 1.4^2=1.96 \\ \hline \dfrac{1}{0.04}=25 & 0.98^2=0.9604 \\ \hline \dfrac{1}{0.06}=16.667& 0.79^2=0.6241 \\ \hline \dfrac{1}{0.08}=12.5& 0.68^2=0.4624\\ \hline \hline \end{array} The figure below shows that the dots are perfectly making a straight line, which means that the author's hypothesis is right. --- b) Now we need to find the constant $c$. From the boxed formula above, the slope is given by $${\rm Slope}=\dfrac{2c}{m}$$ Thus, $$c=\dfrac{m\;{\rm Slope}}{2}$$ Plug the slope from the figure below and the mass from the given information. $$c=\dfrac{50\times 10^{-3}\times 0.04}{2}=1\times 10^{-3}\;\rm kg.m^3/s^2$$ Therefore, $$c=\color{red}{\bf 1\times 10^{-3}}\;\rm J.m$$
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