Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 6 - Review - Page 264: 32

Answer

$$37$$

Work Step by Step

We start with $$k=\frac{1}{2}mv^2$$ First, we multiply both sides by $2$ as follows: $$2k=2\frac{mv^2}{2}\Longrightarrow 2k=mv^2 $$ By dividing by $v^2$, we have $$\frac{2k}{v^2}= \frac{v^2m}{v^2} \Longrightarrow m= \frac{2k}{v^2} .$$ Now, by substitution, we have $$m= \frac{2k}{v^2} =\frac{2*460 }{5^2}=36.8\approx 37$$
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