A skier starts from rest at the top of a hill that is inclined at $10.5$ degrees with the horizontal. The hillside is $200$ m long, and the coefficient of friction between the snow and the skis is $0.075$. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of the snow before coming to rest?
So far, I set the $E_p$ and $E_k$ (potential and kinetic energy) equal to each other since all of the $E_p$ at the top is converted to $E_k$ at the bottom. Using right triangle trig, I found height of the hill to be $36.4$ m. The mass canceled on both sides and I solved for velocity: $V=26.7$ m/s. I thought that the work the snow does on the level portion must be equal to the $E_k$ since the skier comes to a rest. I set $E_k$ equal to the work done by friction, putting the velocity I found in $E_k$: $.5mv^2=coeff.*m*a*d$ The masses cancel, but I am still left with the problem of having both the acceleration and distance in the formula. I don't know what to do from here, and I'm starting to think that maybe this is the incorrect approach to this problem. Can anyone help?
Best Answer
Your approach is right, but you are forgetting a couple things. First off, at the end when you set $\frac{1}{2}mv^2 = f_fd,$ you use the wrong thing for the force of friction. The kinetic friction force is $f_f=\mu N$ where $N$ is the normal force. At the bottom the normal force is just $N = mg,$ so you should set $$\frac{1}{2}mv^2 =\mu mg d. $$
The second thing is you appear to have forgotten the friction on the slope in your energy considerations when computing the speed $v$ at the bottom. The friction is going to slow the skier down, so you're going to need to subtract the energy loss from friction off of the kinetic energy you find at the bottom of the slope (before you set equal to $\frac{1}{2}mv^2$ and solve for $v$). The energy loss due to friction is $f_fL$ where this time $L = 200m.$ The force of friction is again $f_f = \mu N,$ but now since the skier's on an incline $N = mg \cos(\theta).$