Question

What coefficient of friction is required to stop a hockey puck sliding at \(12.5 \mathrm{~m} / \mathrm{s}\) initially over a distance of \(60.5 \mathrm{~m} ?\)

Short answer

Based on the given initial velocity of the hockey puck (\(12.5 \mathrm{~m} / \mathrm{s}\)) and stopping distance (\(60.5 \mathrm{~m}\)), we calculated the acceleration required to stop the puck and then used Newton's laws of motion to relate this acceleration to force and friction. The result is that the coefficient of friction needed to stop the hockey puck is approximately \(0.138\).

Step-by-step solution

List the given information

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 \(v_f = 0\).

Find the acceleration required to stop the puck

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 \(a = \frac{(0)^2 - (12.5)^2}{2(60.5)}\) \(a = -\frac{156.25}{121}\)

Relate acceleration to force and friction

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: \(\mu mg = ma\)

Solve for the coefficient of friction

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 \(0.138\).