Ahockey puck is hit on a frozen lake and starts moving with a speed of 12.1 m/s. exactly 6.10 s later, its speed is 6.10 m/s. the acceleration of gravity is 9.81 m/s 2 .
a) what is its average acceleration?
b) what is the coefficient of kinetic friction between the puck and the ice?

a) a= -0.9836 m/s²

b) μk = 0.1

Explanation:

Hockey puck Kinematics

Because the hockey puck moves with uniformly accelerated movement we apply the following formula:

 Formula (1)

Where:  

a:  average acceleration in m/s²

t :time interval time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

Data

v₀=  12.1 m/s

vf= 6.1 m/s

t= 6.1 s

a)  Calculation of the average acceleration

We apply the formula (1)

a = -0.9836 m/s²

b) Calculation of the coefficient of kinetic friction between the puck and the ice

We apply Newton's second law:

∑F = m*a Formula (2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

g = 9.81 m/s² : acceleration due to gravity

hockey puck weight  (W)

W= m*g = 9.81*m ,   m: hockey puck mass

Friction force between the puck and the ice : Ff

Ff= μk*N   Formula (3)

μk: coefficient of kinetic friction

N : Normal force (N)  

Calculating normal force

We apply the formula (2) for the vertical movement (y):

∑Fy = m*ay     ,  ay=0

N-W=0

N=W= 9.81*m

Calculating Friction force (Ff)

We replace N= 9.81*m  in the formula (2)

Ff= μk*9.81* m

Ff= 9.81* m*μk

Coefficient of kinetic friction between the puck and the ice

We apply the formula (2) for horizontal movement (x)

We choose the positive direction, the direction of the hockey puck 's movement and apply equation (2):

∑Fx = m*ax  , ax= a = -0.9836 m/s²  

-Ff= m*a

-μk*9.81* m = m*(-0.9836 ) : we divide by -m on both sides of the equation

μk = 0.9836/9.81

μk = 0.1