Question

5 a hockey puck slides across the ice with an acceleration of -0.67 m/s². if the puck has a mass of 160 g, what is the coefficient of kinetic friction between the puck and the ice?
μk = 0.62
μk = 0.051
μk = 0.068
μk = 0.11

Explanation:

Step1: Apply Newton's second law

According to Newton's second law \(F = ma\). The frictional force \(F_f=\mu_k N\), and on a horizontal surface \(N = mg\). So \(F_f=\mu_k mg\). Also, \(F = F_f\) (since the only horizontal force is the frictional force causing deceleration), and \(a=- 0.67\ m/s^2\), \(m = 160\ g=0.16\ kg\), \(g = 9.8\ m/s^2\). Then \(\mu_k mg=ma\).

Step2: Solve for \(\mu_k\)

Cancel out \(m\) from both sides of the equation \(\mu_k mg=ma\). We get \(\mu_k=\frac{a}{g}\). Substitute \(a=- 0.67\ m/s^2\) and \(g = 9.8\ m/s^2\) into the formula. \(\mu_k=\frac{\vert - 0.67\vert}{9.8}\approx0.068\)

Answer:

\(\mu_k = 0.068\) (the third option)