A 62.0 skier is moving at 6.10 on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.30 long. The coefficient of kinetic friction between this patch and her skis is 0.330. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 3.05 high.
To answer this question, we need to break it down into smaller parts.
1. Determine the acceleration on the rough patch:
- The formula to calculate the acceleration due to friction is given by: a = μ * g, where μ is the coefficient of kinetic friction and g is the acceleration due to gravity.
- The acceleration will be negative since it is acting opposite to the direction of motion.
- Substituting the values, we get: a = 0.330 * 9.8 = -3.234 m/s² (rounded to three decimal places).
2. Calculate the time taken to cross the rough patch:
- We can use the kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
- The skier's initial velocity (u) is 6.10 m/s, and the acceleration (a) is -3.234 m/s².
- The final velocity (v) can be calculated using: v² = u² + 2as, where s is the distance covered (3.30 m in this case).
- Rearranging the equation, we get: v = sqrt(u² + 2as).
- By substituting the values, we find: v = sqrt(6.10² + 2 * (-3.234) * 3.30) = 5.143 m/s (rounded to three decimal places).
- Plugging the values into the first equation, we can solve for t: 5.143 = 6.10 + (-3.234) * t.
- Solving for t, we find: t ≈ 0.987 seconds (rounded to three decimal places).
3. Calculate the velocity at the bottom of the icy hill:
- Since the hill is frictionless, the skier's acceleration will be equal to the acceleration due to gravity (9.8 m/s²) acting in the downward direction.
- We can use the kinematic equation: v² = u² + 2as, where u is the initial velocity (5.143 m/s), a is the acceleration (9.8 m/s²), and s is the height of the hill (3.05 m).
- Rearranging the equation to solve for v, we find: v = sqrt(u² + 2as).
- Substituting the values, we calculate: v = sqrt(5.143² + 2 * 9.8 * 3.05) = 9.029 m/s (rounded to three decimal places).
Therefore, the skier's velocity at the bottom of the icy hill is 9.029 m/s.