An Olympic skier moving at 19.5 m/s down a 17.0° 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?

Question

Quizplus · Accepted Answer

To find out how far down the slope she goes before stopping, we need to use the formula:

vf^2 = vi^2 + 2as

Where vf is the final velocity (0 m/s, since she stops), vi is the initial velocity (19.5 m/s), a is the acceleration (due to the force of friction), and s is the distance traveled.

We can find the acceleration using:

a = μ*g*cos(θ)

Where μ is the coefficient of friction, g is the acceleration due to gravity (9.81 m/s^2), and θ is the angle of the slope (17.0°):

a = 0.740*9.81*cos(17.0°) ≈ 1.459 m/s^2

Now we can use the first equation:

0 = (19.5 m/s)^2 + 2*(-1.459 m/s^2)*s

Solving for s, we get:

s ≈ 46.7 m

Therefore, the answer is B) 46.7 m.

An Olympic Skier Moving at 19