91Ó°ÊÓ

We are given the initial velocity of the puck, \(v_i = 12.5 \mathrm{~m} / \mathrm{s}\) and the stopping distance, \(d = 60.5 \mathrm{~m}\). The final velocity of the puck when it stops will be \(v_f = 0\).

We can use the following equation to find the acceleration, \(a\): \(v_f^2 = v_i^2 + 2ad\) Rearranging the equation to solve for 'a', we get \(a = \frac{v_f^2 - v_i^2}{2d}\) Substituting the values, we get \(a = \frac{(0)^2 - (12.5)^2}{2(60.5)}\) \(a = -\frac{156.25}{121}\)

According to Newton's second law of motion, force (\(F\)) equals mass (\(m\)) times acceleration (\(a\)): \(F = ma\) The force of friction (\(F_f\)), in this case, acts opposite to the motion of the puck and is responsible for stopping it. Therefore, the force of friction can be expressed as \(F_f = \mu N\) Where \(\mu\) is the coefficient of friction, and \(N\) is the normal force acting on the puck. Since the puck is sliding on a flat surface, the normal force equals the gravitational force acting on the puck, \(N = mg\). Thus, the force of friction equation becomes \(F_f = \mu mg\) Since the force of friction is responsible for stopping the puck, we can equate it with the force derived from Newton's second law: \(\mu mg = ma\)

Now, we can solve for \(\mu\) from the equation \(\mu mg = ma\). The mass \(m\) can be canceled out: \(\mu = \frac{a}{g}\) Where \(g\) is the acceleration due to gravity, which is approximately \(9.81 \mathrm{~m} / \mathrm{s}^2\). Substituting the values for \(a\) and \(g\), we get \(\mu = \frac{-\frac{156.25}{121}}{9.81}\) \(\mu \approx 0.138\) Therefore, the coefficient of friction required to stop the hockey puck sliding at \(12.5 \mathrm{~m} / \mathrm{s}\) initially over a distance of \(60.5 \mathrm{~m}\) is approximately \(0.138\).