A0.50 kg puck is sliding on a horizontal shuffleboard court is slowed to rest by a frictional force of 1.2 newtons. what is the coefficient of kinetic friction between the puck and the surface of the shuffleboard court?

-The u is what you are trying to find, also known as coefficient of friction.
-Fn is your mass in Newton's, which is 0.5 x 9.8 which equals 4.9
-Fmax is your frictional force. 

When you solve for , your equation becomes
 
/ Fn =

  = 1.2N/4.9N 

(coefficient of friction) = 0.25

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