Answer:
Explanation:
Given :
Initial speed [tex]v_{o} = 9.5 \frac{m}{s}[/tex]
Displacement [tex]x = 37.4[/tex] m
From the kinematics equation,
[tex]v^{2} - v^{2} _{o} = 2ax[/tex]
Where [tex]v^{2} =[/tex] final velocity, in our example it is zero ([tex]v =0[/tex]), [tex]a =[/tex] acceleration.
[tex]a =- \frac{90.25}{ 2 \times 37.4}[/tex]
[tex]a =- 1.21 \frac{m}{s^{2} }[/tex]
From the formula of friction,
[tex]F =- \mu _{k } N[/tex]
Minus sign represent friction is oppose the motion
Where [tex]N = mg[/tex] ( normal reaction force )
[tex]ma = -\mu _{k} m g[/tex] ( ∵ [tex]g = 9.8 \frac{m}{s^{2} }[/tex] )
So coefficient of friction,
[tex]\mu_{k} = \frac{1.21}{9.8}[/tex]
[tex]\mu_{k} = 0.12[/tex]
Therefore, the coefficient of kinetic friction between the puck and the ice is [tex]\mu _{k} =[/tex] 0.12 .
