College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 11 - Problems - Page 432: 68


(a) $\sqrt{\frac{F~L}{m}}$ has units of $m/s$ which are units of speed. (b) $v = C~\sqrt{\frac{F~L}{m}}$, where $C$ is a dimensionless constant

Work Step by Step

$F$ has units of $N$ which can also be expressed as $kg~m~s^{-2}$ $L$ has units of $m$ $m$ has units of $kg$ We can find the units of $\sqrt{\frac{F~L}{m}}$: $\sqrt{\frac{(kg~m~s^{-2})(m)}{kg}} = \sqrt{m^2~s^{-2}} = m/s$ $\sqrt{\frac{F~L}{m}}$ has units of $m/s$ which are units of speed. (b) Let's assume that $~C~F^a~L^b~m^c = v~$, where $C$ is a dimensionless constant. Then: $(kg~m~s^{-2})^a~m^b~kg^c = m~s^{-1}$ We can consider the units of seconds: $(s^{-2})^a = s^{-1}$ $-2a = -1$ $a = \frac{1}{2}$ We can consider the units of $m$: $m^a~m^b = m^1$ $a+b = 1$ $b = 1-a$ $b = 1-\frac{1}{2}$ $b = \frac{1}{2}$ We can consider units of $kg$: $kg^a~kg^c = kg^0$ $a+c = 0$ $c = -a$ $c = -\frac{1}{2}$ We can write the equation with the exponents: $v = C~F^a~L^b~m^c$ $v = C~F^{1/2}~L^{1/2}~m^{-1/2}$ $v = C~\sqrt{\frac{F~L}{m}}$
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