# Homework Help: Ski Slope question

1. Apr 10, 2004

### st3dent

2. Two mountains are at elevations of 850m and 750m above the valley between them. A ski run extends from the top of the higher peak to the top of the lower one, with a total length of 3.2km and an average slope of 30 degrees,
a) A skier starts from rest on the higher peak. At what speed will be arrive at the top of the lower peak if he just coasts without using poles?
b)How large a coefficient of friction would be tolerable without preventing him from reaching the lower peak?

what i've done:
2a) (m)(9.8)(850) = (m)(9.8)(750) + (0.5)(m)($$vf^2$$)
$$vf^2$$ = 1960
$$vf$$ = 44.27 m/s

b)I am stuck on this one.

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2. Apr 10, 2004

### HallsofIvy

Assume that the skier just arrives at the second peak with 0 velocity. How much energy must he/she have lost? Okay, now use "energy = force*distance" to calculate the friction force necessary to account for that energy loss.

3. Apr 10, 2004

### Chen

(a) looks right, I didn't check your calculation but the formula is correct.

As for (b), let's write down the work-energy equation:

$$W_{friction} = \Delta E_m = \Delta E_p + \Delta E_k$$

Our worst case scenario would be if the coefficient of friction would be so large, that the skier would reach the lower peak just barely - without any speed. In that case his kinetic energy would be 0 at first (since he starts from rest) and 0 at the end (since he has no speed when he reaches the peak):

$$W_f = (mgh_2 - mgh_1) + (0 - 0) = mg(750m - 850m) = -100mg$$

Now what is the work of friction? It is the force of friction times the distance, which is 3.2km or 3,200 meters. Don't forget the work is negative, though, because the force vector is directly opposite to the distance vector all the time:

$$-xf_k = -3200f_k = -100mg$$

$$f_k = N\mu _k = \frac{100mg}{3200} = \frac{mg}{32}$$

$$\mu _k = \frac{mg}{32N}$$

Now all you have to do is find the normal force, N...

4. Apr 10, 2004

### st3dent

I understand the following:

Since the change in kinetic energy is zero,
Ei = Ef
mgh1 = mgh2 + Ffrd
mgh1 - mgh2 = Ffrd
mg(750 - 850) = Ffrd

Since Ffrd = $$\mu _k$$Fnd

Fnd = mg(750 - 850)
Fn = -100mg / d
Fn = -100mg/-3200
Fn = mg/32

AS;
Ek = 0.5mv2

In part a), i found the speed of the skier at the top of the second slope to be 44.27 m/s.
Then,
(m)(9.8)(850) = (m)(9.8)(750) + (0.5)(m)(v2)
(m)(9.8)(850) = (m)(9.8)(750) + (0.5)(m)(44.272)
However, i can't solve for m, since it cancels out.

I do know that the force down the slop is mgsin30 and that
Fn, the normal force, is mgcos30. But how can i use this to solve this problem?

Last edited: Apr 10, 2004
5. Apr 10, 2004

### st3dent

I THINK I have a solution now.

Since Ffrd = mg / 32
and Ffr = $$\mu _k$$mgcos30

then $$\mu _k$$mgcos30 = mg/32
$$\mu _k$$ = mg/32mgcos30
$$\mu _k$$ = 1/32cos30
$$\mu _k$$ = 0.036

6. Apr 11, 2004

Yes, 0.036.