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16 January, 15:02

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?

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  1. 16 January, 15:25
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    -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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