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.
  
  ${F}_{r} \cdot d = \frac{1}{2} m {v}^{2}$
  
  $d = \frac{\frac{1}{2} m {v}^{2}}{5 m} = {v}^{2} / 10 = {12}^{2} / 10 = \frac{144}{10} = 14.4 m$
explanation: The document contains a physics problem that involves calculating the distance an object will continue to move given its initial velocity and the kinetic friction coefficient. The text includes both the question and the detailed solution, leading to the answer that the distance is 14.4 meters.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 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.
  
  ${F}_{r} \cdot d = \frac{1}{2} m {v}^{2}$
  
  $d = \frac{\frac{1}{2} m {v}^{2}}{5 m} = {v}^{2} / 10 = {12}^{2} / 10 = \frac{144}{10} = 14.4 m$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a physics problem that involves calculating the distance an object will continue to move given its initial velocity and the kinetic friction coefficient. The text includes both the question and the detailed solution, leading to the answer that the distance is 14.4 meters.
  
  Label: question-and-answer