text: If an object is moving at 12 m/s over a surface with a kinetic friction coefficient of u_k=5 /g, how far will the object continue to move?
  
  Apr 2, 2017
  
  The distance is $= 14.4 m$
  
  Explanation:
  
  Let the mass of the object be $= m k g$
  
  $v = 12 m {s}^{-} 1$
  
  The coefficient of kinetic friction is
  
  ${\mu}_{k} = {F}_{r} / N$
  
  $N = m g$
  
  So,
  
  ${F}_{r} = {\mu}_{k} \cdot N$
  
  $= \frac{5}{g} \cdot m g = 5 m N$
  
  The work done by ${F}_{r}$ is equal to the kinetic energy of the object.
  
  ${F}_{r} \cdot d = \frac{1}{2} m {v}^{2}$
  
  $d = \frac{\frac{1}{2} m {v}^{2}}{5 m} = {v}^{2} / 10 = {12}^{2} / 10 = \frac{144}{10} = 14.4 m$