A baseball player slides to a stop on level ground with an initial speed of 6.00 m/s and mass 65.0 kg, experiencing a constant friction force of 450 N. Using energy considerations, how far does he slide before stopping?

1. On level ground, friction alone does negative work to remove all of the player's kinetic energy, so f d equals one half m vi squared. 2. Solving for the distance, d = m vi squared divided by (2 f). 3. Substituting the values, d = (65.0 kg)(6.00 m/s)^2 divided by (2 times 450 N). 4. The numerator is 65.0 times 36.0, which is 2340 kg*m^2/s^2, and the denominator is 900 N. 5. Dividing gives d = 2.60 m. 6. 5.20 m comes from forgetting the factor of 2 in the denominator, and 1.30 m and 0.867 m come from dividing by an extra factor of 2 or 3 beyond the correct denominator. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.5 Nonconservative Forces_