A 0.50 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