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

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snehashish65 snehashish65 · Answer 2 · 2021-12-04T10:04:28+01:00

The coefficient of kinetic friction between the puck and the surface of the shuffleboard court is 0.24.

Given data:

The mass of puck is, m = 0.50 kg.

The magnitude of frictional force is, f = 1.2 N.

The force which tends to oppose the motion between an object and the surface is known as frictional force. And its value is,

[tex]f = \mu \times mg[/tex]

here, [tex]\mu[/tex] is the coefficient of kinetic friction between the puck and the surface of the shuffleboard court.

Solving as,

[tex]1.2 = \mu \times 0.50 \times 9.8\\\\\mu = \dfrac{1.2}{0.50 \times 9.8}\\\\\mu = 0.24[/tex]

Thus, we can conclude that the coefficient of kinetic friction between the puck and the surface of the shuffleboard court is 0.24.

Learn more about the kinetic friction here:

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