Question

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

Answer 1

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

Step-by-step explanation:

To find the coefficient of kinetic friction between the puck and the surface of the shuffleboard court, we can use the equation:

frictional force = coefficient of kinetic friction × normal force

Since the puck is sliding horizontally, the normal force is equal to the weight of the puck, which is given by:

weight = mass × gravitational acceleration

Now we can substitute the given values into the equation and solve for the coefficient of kinetic friction:

1.2 N = coefficient of kinetic friction × (0.50 kg × 9.8 m/s^2)

coefficient of kinetic friction = 1.2 N / (0.50 kg × 9.8 m/s^2) = 0.24

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

Answer 2

-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