6. a skier starts from rest at the top of a frictionless incline of height 20.0 m. at the bottom of the…

Explanation:

Step1: Use conservation of mechanical energy for part (a)

The initial potential - energy at the top of the incline is converted into kinetic energy at the bottom. The initial potential energy of the skier is $U = mgh$, and the kinetic energy at the bottom is $K=\frac{1}{2}mv^{2}$. By the conservation of mechanical energy $U = K$, so $mgh=\frac{1}{2}mv^{2}$. Canceling out the mass $m$ on both sides, we get $v=\sqrt{2gh}$. Given $h = 20.0\ m$ and $g = 9.8\ m/s^{2}$, then $v=\sqrt{2\times9.8\times20.0}=\sqrt{392}\approx 19.8\ m/s$.

Step2: Analyze the forces and use kinematic equations for part (b)

The frictional force on the horizontal surface is $F_f=\mu_k N$, where $N = mg$ (since the normal force equals the weight on a horizontal surface). So $F_f=\mu_kmg$. According to Newton's second - law $F = ma$, and $F_f=ma$, then $\mu_kmg=ma$, and the acceleration $a=-\mu_kg$. The kinematic equation $v_f^{2}-v_i^{2}=2ax$ is used, where $v_f = 0$ (the skier comes to rest), $v_i$ is the speed at the bottom of the incline, and $a=-\mu_kg$. We know $v_i\approx19.8\ m/s$, $\mu_k = 0.210$, and $g = 9.8\ m/s^{2}$. From $v_f^{2}-v_i^{2}=2ax$, we can solve for $x$: [ \begin{align*} 0 - v_i^{2}&=2(-\mu_kg)x\ x&=\frac{v_i^{2}}{2\mu_kg} \end{align*} ] Substitute $v_i\approx19.8\ m/s$, $\mu_k = 0.210$, and $g = 9.8\ m/s^{2}$ into the formula: [ \begin{align*} x&=\frac{(19.8)^{2}}{2\times0.210\times9.8}\ &=\frac{392.04}{4.116}\ &\approx95.2\ m \end{align*} ]

Answer:

(a) The skier's speed at the bottom is approximately $19.8\ m/s$. (b) The skier travels approximately $95.2\ m$ on the horizontal surface before coming to rest.