Question

The coefficient of kinetic friction between a 0.165 kg hockey puck and ice is μk = 0.010. How far will the puck travel if it is hit with an initial velocity of 4.266 m/s?

Answer 1

they give mass m=0.165 kg, kinetic friction coefficient uk=0.010, and initial velocity at 4.266 m/s.

You can do a force sum for the force diagram above. The vertical is balanced so no acceleration, the horizontal has a net force from friction which causes acceleration.
\[\sum F _{x}=N-W=0\]
\[\sum F _{y}=-F _{f}=m*a\]

Answer 2

Friction force is the coefficient times the normal force on the object. The weight W is mass times gravity

From first equation
N = W =m * g
and second equation
\[-\mu _{k}*N=m*a\]

\[a=\frac{ -\mu _{k}*m*g }{ m }= -\mu _{k}*g\]
a=-0.010*-9.8 = 0.098 m/s^2

Answer 3

Starting at 4.266 m/s initially, and ending with zero velocity,and having a constant acceleration you can use the kinematic equation

\[\large v _{f}^2=v _{i}^2+2*a*d\]
0 = 4.266^2 - 2* 0.098 *d
d=92.9 m