Lascano15
Answered

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A 0.50-kilogram puck 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?
(1) 0.24 (3) 0.60
(2) 0.42 (4) 4.1

Sagot :

sbcardinals


-The u is what you are trying to find, also known as coefficient of friction.
-Fn is your mass in Newtons, 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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