Answer:
[tex]\mu_{k} = 0.034[/tex]
Explanation:
Let assume that hockey puck is moving on a horizontal ground. The corresponding equations of equilibrium for the hockey puck are:
[tex]\Sigma F_{x} = -\mu_{k}\cdot N = m\cdot a\\\Sigma F_{y} = N - m\cdot g = 0[/tex]
The following expression is built after some algebraic handling:
[tex]-\mu_{k}\cdot m\cdot g = m\cdot a[/tex]
The deceleration experimented by the object is:
[tex]a = \frac{1\,\frac{m}{s}-2\,\frac{m}{s}}{3\,s}[/tex]
[tex]a = -0.333\,\frac{m}{s^{2}}[/tex]
The kinetic coefficient of friction is:
[tex]\mu_{k} = -\frac{a}{g}[/tex]
[tex]\mu_{k} = - \frac{(-0.333\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{k} = 0.034[/tex]
