91Ó°ÊÓ

In order to find the acceleration, one can use the kinematic equation \(v_f^2 = v_i^2 + 2ad\), where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(d\) is the distance. After inserting the given values where the final velocity \(v_f\) is 0 m/s (since the puck comes to rest), the initial velocity \(v_i\) is 14 m/s, and the distance \(d\) is 56 m, this gives us \(0^2 = 14^2 + 2a \times 56\). Solve this equation for \(a\) (acceleration).

The frictional force \(f_k\) can be equated to the product of the coefficient of kinetic friction \(\mu_k\) and the normal force \(N\) (which in this case is simply the weight of the object, as the puck is sliding on a level friction), \(f_k = \mu_k \cdot N\). However, as there's no external force acting horizontally, the frictional force also equates to the product of mass \(m\) and acceleration \(a\), \(f_k = m \cdot a\). As we're asked to find the coefficient of friction and not the friction itself, we can eliminate \(f_k\) and \(m\) from the equations to get \(\mu_k = a / g\), where \(g\) is the acceleration due to gravity. Here, using \(a\) calculated in Step 1 and \(g = 9.8 \, m/s^2\), we can find the value of \(\mu_k\).