Question

consult multiple - concept example 5 for insight into solving this problem. a skier slides horizontally along the snow for a distance of 11.3 m before coming to rest. the coefficient of kinetic friction between the skier and the snow is 0.0576. initially, how fast was the skier going?

Explanation:

Step1: Find the acceleration

According to Newton's second law \(F = ma\), and the frictional force \(F_f=\mu_k N\). Since \(N = mg\) (normal force equals weight), then \(F_f=\mu_k mg\). So \(ma=-\mu_k mg\) (negative because it's decelerating), and \(a =-\mu_k g\). Substituting \(\mu_k = 0.0576\) and \(g = 9.8\ m/s^2\), we get \(a=- 0.0576\times9.8=-0.56448\ m/s^2\).

Step2: Use the kinematic equation

The kinematic equation \(v^2=v_0^2 + 2ax\). Since \(v = 0\) (comes to rest), \(x = 11.3\ m\), and \(a=-0.56448\ m/s^2\). Rearranging for \(v_0\), we have \(v_0=\sqrt{- 2ax}\). Substituting the values: \(v_0=\sqrt{-2\times(- 0.56448)\times11.3}\).

$$v_0=\sqrt{2\times0.56448\times11.3}=\sqrt{12.777}=3.57\ m/s$$

Answer:

\(3.57\ m/s\)