Question

consult multiple-concept example 5 for insight into solving this problem. a skier slides horizontally along the snow for a distance of 28.0 m before coming to rest. the coefficient of kinetic friction between the skier and the snow is 0.0555. initially, how fast was the skier going?

Explanation:

Step1: Find the acceleration of the skier

According to Newton's second law \(F = ma\). The frictional force \(F_f=\mu_kmg\), so \(ma =-\mu_kmg\) (negative because it's decelerating). Then \(a=-\mu_kg\). Substituting \(\mu_k = 0.0555\) and \(g = 9.8\ m/s^2\), we get \(a=- 0.0555\times9.8=-0.5439\ m/s^2\).

Step2: Use the kinematic equation \(v^2 = v_0^2+2ax\)

We know that \(v = 0\) (comes to rest), \(x = 28.0\ m\), and \(a=-0.5439\ m/s^2\). Rearranging the equation for \(v_0\), we have \(v_0=\sqrt{v^2 - 2ax}\). Substituting the values: \(v_0=\sqrt{0-2\times(- 0.5439)\times28.0}\).

$$v_0=\sqrt{2\times0.5439\times28.0}=\sqrt{30.4584}\approx5.52\ m/s$$

Answer:

\(5.52\ m/s\)