## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 5 - Section 5.1 - Energy and Energy Changes - Checkpoint - Page 191: 5.1.1

a) 65 J

#### Work Step by Step

m= 5.0 kg v = 26 m/s $E_{k}$= $\frac{1}{2}mv^{2} = \frac{1}{2}\times5.0kg\times(26m/s)^{2}$ = 65 J.

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