Friction and motion...?

Question

I am really confused on this question. Could someone please assist me? Thank you!

An Olympic skier moving at 11 m/s down a 18 degree slope encounters a region of wet snow of coefficient of kinetic friction of mu = .36. The acceleration of gravity is 9.8 m/s^2. How far down the slope does she travel before coming to a halt? Answer in units of m.

Answer 1

u = sqrt(2aS); where u = 11 mps, a = f/m = [W sin(theta) - k W cos(theta)]/m = mg[sin(theta) - k cos(theta)]/m = g[sin(theta) - k cos(theta)]. Theta = 18 deg slope, k = mu = .36

Thus, S = u^2/2a, the stoppage distance S along the slope if a < 0 a deceleration. You can find a from a = g[sin(theta) - k cos(theta)], which was derived above.

You can do the math. Study the physics though to understand what is happening. That is, the frictional force is dragging back on the portion of the skier's weight that is aimed down the slope. When that frictional force exceeds the slope oriented weight, there is a deceleration. That's why a must be negative, to show the deceleration.