An Olympic skier moving at 20.0 m/s down a 30.0 degrees slope encounters a region of wet snow, of coefficient of friction uk = 0.740. How far down the slope does she go before stopping?

To calculate how far the Olympic skier goes down the slope before stopping, we can use the concept of friction and the equations of motion.

The force of friction between the skier and the wet snow can be calculated using the equation:

frictional force (Ff) = coefficient of friction (uk) * normal force (N)

The normal force (N) can be calculated using the equation:

normal force (N) = mass (m) * gravitational acceleration (g) * cos(angle)

The gravitational acceleration (g) can be assumed to be approximately 9.8 m/s^2.

Once we have the frictional force (Ff), we can calculate the acceleration (a) using Newton's second law of motion:

Ff = m * a

Finally, we can use the equations of motion to calculate the distance (d) traveled by the skier:

v^2 = v0^2 + 2 * a * d

where v is the final velocity (0 m/s in this case since the skier stops) and v0 is the initial velocity (20.0 m/s).

Let's now calculate the solution step by step:

Step 1: Calculate the normal force (N):
N = m * g * cos(angle)
= m * (9.8 m/s^2) * cos(30 degrees)

Step 2: Calculate the frictional force (Ff):
Ff = uk * N

Step 3: Calculate the acceleration (a):
Ff = m * a

Step 4: Calculate the distance (d):
v^2 = v0^2 + 2 * a * d
0^2 = (20.0 m/s)^2 + 2 * a * d

Since the skier eventually comes to a stop, the final velocity (v) is 0 m/s. Solving for d:

d = -(v0^2) / (2 * a)

Now, let's plug in the given values and solve the equations to find the answer.

145m