Physics: Coefficient of Kinetic Friction

Question

A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. Exactly 5.0s later, its speed is 6.0 m/s.

What is the coefficient of kinetic friction between the puck and the ice?

Please show clear steps and explanations. Thanks.

Answer 1

From Newton's 2nd Law of Motion,

F = ma

where

F = frictional force acting on the puck

m = mass of the puck

a = acceleration of the puck

a = (6 - 12)/5 = -6/5 = - 1.2 m/sec^2

F = coeff of friction * normal force

F = coeff of friction * mg

where

m = mass of the puck (same as above description)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Therefore,

-(coeff of friction) * (m) * 9.8 = m (-1.2)

NOTE the negative sign attached to F = -coeff of friction*m*9.8. This simply denotes that the frictional force is acting opposite that of the puck's direction.

NOTE as well that since "m" appears on both sides of the equation, then it will simply cancel out, hence

coeff of friction * 9.8 = 1.2

coeff of friction = 1.2/9.8 = 0.1224

Answer 2

initial speed is 12 m/s

final speed is 6 m/s

time elapsed is 5 s

acceleration = change in v / change in t = 12 - 6 / 5 = 1.2 m/s^2

frictional force = mass * acceleration (Newton's second law)

normal force = weight (no other forces in the Y direction draw an FBD if it helps and no acceleration in Y) = mass *gravity

frictional force = normal force * coefficient = mass * gravity * coefficient

set the two equal to each other

Masses cancel and solve for coefficient (sorry I'm not sure how to insert greek letters)

coefficient = (6 / (5*9.8)) = 1.2

Hope that helps