# Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 33

$\mu_k = 0.037$

#### Work Step by Step

The work done by friction is equal in magnitude to the initial kinetic energy of the suitcase; $W_f = KE$ $F_f~d = \frac{1}{2}mv^2$ $mg~\mu_k~d = \frac{1}{2}mv^2$ $\mu_k = \frac{v^2}{2~g~d}$ $\mu_k = \frac{(1.2~m/s)^2}{(2)(9.80~m/s^2)(2.0~m)}$ $\mu_k = 0.037$

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