An Olympic skier moving at 22.0 m/s down a 30.0° slope encounters a region of wet snow,of coefficient of friction µ_k = 0.740.How far down the slope does she go before stopping? A) 119 m B) 145 m C) 175 m D) 199 m

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Quizplus · Accepted Answer

First, we need to find the component of gravity acting parallel to the slope. F_parallel = m * g * sin(theta) = (m * 9.8 * sin(30°)) Next, we need to find the force of friction: F_friction = µ_k * F_N We know that F_N = m * g * cos(theta) = (m * 9.8 * cos(30°)) Therefore, F_friction = 0.740 * (m * 9.8 * cos(30°)) Since the skier stops, we know that: F_parallel = F_friction Therefore, (m * 9.8 * sin(30°)) = 0.740 * (m * 9.8 * cos(30°)) Solving for m, we find that m cancels out of the equation: tan(30°) = 0.740, which gives us: m = (22.0^2) / (2 * 9.8 * sin(30°) * 0.740) Using the equation for distance: d = (v^2) / (2 * µ * g), We can plug in the values we know to get: d = (22.0^2) / (2 * 0.740 * 9.8 * cos(30°)) = 175 meters. Therefore, the answer is C.

An Olympic Skier Moving at 22