Question

A skier with a mass of 90 kg starts from rest down a hill inclined at 17◦. He skis 100 m down the hill and then coasts for 70 m along level snow until he stops. Find the coefficient of kinetic friction between the skis and the snow. What velocity does the skier have at the bottom of the hill?


here where I'm at so far, I found that Fn-mgcos17=0, Fn=mgcos17, Fn=843.5 N

I also found that Ff=mgsin17-Fskier

I know that uk= Ff/Fn but I have no idea how to find Ff

Answer 1

Seeing how nobody has yet taken this task up I'll give it a try. The main problem you have is that you just assume that all the forces are in equilibrium and that there is no total net force acting on the skier.

But the task clearly states that the skier went from zero speed and skied to the bottom of the cliff at which point he coasted to a full stop. This implies that the skier **accelerated** to some final speed at the end of the downhill and then **decelerated** to a full stop. ANY acceleration immediately has to involve some kind of net force acting on the body or else we would be braking one of Newtons laws.



So these are the forces acting on the skier in the two situations, on a slope and on a horizontal ground. Ff is the force of friction, Fg is the force from gravitational attraction, Fgf is the forward component of that force and N is the "normal" force or the component of the force felt from gravitational attraction perpendicular to the surface. We need this force because that's the "weight" the force of friction "feels".

$$
F_{\text{total1}} = F_{gf} - F_f \\
F_{\text{total2}} = F_f\\
------------------------------------\\
ma_1 = F_g\sin\alpha - \mu N\\
ma_2 = \mu F_g\\
------------------------------------\\
ma_1 = F_g\sin\alpha - \mu F_g \cos\alpha\\
ma_2 = \mu mg\\
------------------------------------\\
ma_1 = mg\sin\alpha - \mu mg\cos\alpha \,\,\, \large{/} :m\\
ma_2 = \mu mg\,\,\, \large{/} :m\\
------------------------------------\\
a_1 = g(\sin\alpha - \mu \cos\alpha)\\
a_2 = g\mu
$$
This is where the first issue appears, we don't know the values of the accelerations in the two cases a1 and a2. We also don't know the value of mu, the coefficient of friction. The only way forward from this is to reduce the number of unknowns. At some point the skier will be crossing from the slope into horizontal flat ground. At some tiny minute moment when that happens the two accelerations should be the same. Mind you, I'm not confident in making this statement, it just feels rather silly to ignore acceleration and assume it's a constant, but at the same time assuming it changes with time and that we can pick a certain moment when they're the same. But whatever, it's the only thing we can do right now.
$$
\mu g = g(\sin\alpha - \mu \cos\alpha)\\
\mu = \sin\alpha - \mu \cos\alpha\\
\mu + \mu \cos\alpha= \sin\alpha \\
\mu = \frac{\sin\alpha}{\cos\alpha +1}\\
\mu = 0.1494479482
$$
Ok neat. Now we also have to get the velocity at the end of the slope which we can do from:
$$v^2 = v_0^2 + 2a[s-s_0)$$
by putting in the initial speed to be zero. The initial position s0 we'll also say to be zero and we know the slope is 100m long. We need to calculate the acceleration which we can do from the forces acting on the body for situation 1, when he's on the slope:
$$a_1 = g(\sin\alpha - \mu \cos\alpha)\\
= 9.81*(\sin 0.29671 - 0.15 \cos 0.29671)\\
= 1.461\frac{m}{s^2}$$
Back into speed:
$$v^2 = v_0^2 + 2a[s-s_0) = 2*1.461*100\dots \\
v = 17.1\frac{m}{s^2}$$
Which is about 60 kph (kilometers per hour). Ok so does any of this make any actual sense in the real world? Well so-so. The measured friction coefficients for snow vary from 0.03 to 1.76 so we're inside the range which is good. Average skiers don't go that fast but apparently this is the on the lower side of downhill racing skiers which can go much much faster, up to 150kph. Faster still are the speed skiers that can go up to 250kph. So it makes sense if we assume that this is a sporty guy/gal skiing rather fast down the hill.

http://hypertextbook.com/facts/2007/TabraizRasul.shtml
http://skiing.about.com/od/skiingtip1/qt/How-Fast-Do-Skiers-Go.htm