10. A skier has just begun descending a 20^circ slope. Assuming that the coefficient of kinctic friction is 0.10 calculate a) the acceleration of the skier. b) his final velocity after 8s.

**a) Calculating the acceleration:**<br /><br />1. **Identify the forces:** The forces acting on the skier are gravity (mg), the normal force (N) perpendicular to the slope, and kinetic friction (f_k) opposing the motion down the slope.<br /><br />2. **Resolve gravity:** The component of gravity parallel to the slope (mg sin θ) accelerates the skier downwards. The component perpendicular to the slope (mg cos θ) is balanced by the normal force.<br /><br />3. **Calculate the frictional force:** The kinetic frictional force is given by f_k = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force. Since N = mg cos θ, we have f_k = μ_k * mg cos θ.<br /><br />4. **Apply Newton's second law:** The net force acting on the skier down the slope is mg sin θ - f_k. According to Newton's second law (F=ma), this net force equals the skier's mass times acceleration (ma).<br /><br />5. **Solve for acceleration:**<br /><br />* mg sin θ - μ_k * mg cos θ = ma<br />* a = g(sin θ - μ_k cos θ) (mass cancels out)<br />* a = 9.8 m/s²(sin 20° - 0.10 * cos 20°)<br />* a ≈ 9.8 m/s²(0.342 - 0.10 * 0.940)<br />* a ≈ 9.8 m/s²(0.342 - 0.094)<br />* a ≈ 9.8 m/s² * 0.248<br />* a ≈ 2.43 m/s²<br /><br />**b) Calculating the final velocity:**<br /><br />1. **Use the equation of motion:** Assuming the skier starts from rest (initial velocity u = 0), we can use the following equation of motion:<br /><br /> v = u + at<br /><br /> where:<br /> * v = final velocity<br /> * u = initial velocity (0 m/s)<br /> * a = acceleration (calculated in part a)<br /> * t = time (8 s)<br /><br />2. **Solve for final velocity:**<br /><br />* v = 0 + (2.43 m/s²)(8 s)<br />* v ≈ 19.4 m/s<br /><br />**Therefore:**<br /><br />* **a) The acceleration of the skier is approximately 2.43 m/s².**<br />* **b) The skier's final velocity after 8 seconds is approximately 19.4 m/s.**<br />