## Precalculus (6th Edition) Blitzer

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