Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 425: 34

Answer

A mass $m=4\text{ gm}$ and velocity $v=6\text{ centimeters per second}$ has a kinetic energy of 72 erg.

Work Step by Step

According to the question, $e\propto m{{v}^{2}}$ which can be written as $e=km{{v}^{2}}$ where $e$ is Kinetic energy, $m$ is the mass, $v$ is the velocity and $k$ is the constant. We have to find the value of constant $k$ for the given data i.e. $m=4,v=6$ and $e=36$. Substitute the values of e, m, and v in the above formula to find the value of k. $\begin{align} & e=km{{v}^{2}} \\ & 36=k\left( 8 \right){{\left( 3 \right)}^{2}} \\ & 36=k\left( 8 \right)\left( 9 \right) \\ & k=\frac{36}{72} \end{align}$ So, $\begin{align} & k=\frac{1}{2} \\ & k=0.5 \end{align}$ Now, calculate the value of e. $\begin{align} & e=0.5m{{v}^{2}} \\ & e=0.5\left( 4 \right){{\left( 6 \right)}^{2}}\ \left( \because \ m=4\And v=6 \right) \\ & e=0.5\left( 4 \right)\left( 36 \right) \\ & e=72\text{ergs} \end{align}$ Therefore, mass $m=4\text{ gms}$ and velocity $v=6\text{ centimetres per second}$ has a kinetic energy of $72\text{ }erg$.
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