Answer:
(a) Ff = 0.128 N 
(b  μk = 0.1135
Explanation:
kinematic analysis
Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:
vf=v₀+a*t Formula (1)
d= v₀t+ (1/2)*a*t² Formula (2)
Where:  
d:displacement in meters (m)  
t : time in seconds (s)
v₀: initial speed in m/s  
vf: final speed in m/s  
a: acceleration in m/s
Calculation of the acceleration of the  hockey puck 
We apply the Formula (1)
vf=v₀+a*t      v₀=5.8 m/s ,  vf=0
0=5.8+a*t
-5.8 = a*t
a= -5.8/t   Equation (1)
We replace a= -5.8/t in the Formula (2)
d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s 
15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  
15.1= 5.8*t-2.9*t
15.1= 2.9*t
t = 15.1 / 2.9
t= 5.2 s
We replace t= 5.2 s in the equation (1)
a= -5.8/5.2
a= -1.115 m/s²
(a) Calculation of the  frictional force (Ff)
We apply Newton's second law
∑F = m*a    Formula (3)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Look at the free body diagram of the  hockey puck in the attached graphic
∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²
-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation
Ff = 0.128 N 
(b) Calculation of the coefficient of friction (μk)
N: Normal Force (N)
W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight 
g: acceleration due to gravity =9.8 m/s² 
∑Fy = 0
N-W=0
N = W
N =  1.127 N
μk = Ff/N
μk = 0.128/1.127
μk = 0.1135