# Homework Help: Skiers - Who Needs 'em?

1. May 23, 2004

### Klassic

I understand that it is required that a user have attempted the problems before posting as this is not a place where people will do your homework for you. I'm a relatively good physics student, and was just handed a 30 problem packet to complete. I've spent the majority of my weekend figuring these problems out, and I'm left with 2 problems that I'm having a great deal of trouble with. The problems are:

A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill. The crest of the second hill is circular with a radius of r=36 m. Neglect friction and air resistance. What is the height of the first hill such that the skier just loses contact with the snow at the crest of the second hill?

I've figured out that to figure out this problem, you must have a strong bearing on the concepts of centripetal acceleration/force. I don't entirely grasp those concepts. I'm having a hard time even getting a grasp on this problem.

Second:

An extreme skier, starting from rest, coasts down a mountain that makes an angle of 25.0 degrees with the horizontal. THe coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she traveling just before she lands?

Again, I'm aware this is not really a "homework help" place. But please realize that I have attempted both problems - and failed miserably. The other problems have obviously tired my brain. ANY help would be very much appreciated. Please~. :)

2. May 23, 2004

### arildno

In the first case, gravity is evidently the force providing centripetal acceleration.
The skier will lose contact with the surface if the NORMAL FORCE IS ZERO.
This means that the force of gravity must provide exactly the centripetal acceleration
(otherwise, there would have been a non-zero normal force)

Hence, the requirement for loss of contact is:
$$mg=m\frac{V^{2}}{r}$$
(I assume you know what numbers to plug in here.. )

3. May 23, 2004

### baffledMatt

What you are looking for is to find the velocity he requires such that at the top of the second hill his centrifugal force (something like $$m v^2 / r$$ where r is the radius of the hill) is no longer balanced by gravity, $$mg$$. Since there is no friction you should be able to calculate his velocity at any time by equating KE to PE, ie

$$1/2 m v^2 = mg \Delta h$$

where $$\Delta h$$ is the difference between his current and initial height.

Hope that helps.

Matt

4. May 23, 2004

### arildno

In the second problem, it is important to separate the motion in 2 distinct parts.
1. Part: Down the slope.
2. Part: Free fall.

The 2.part is rather simple, so I'll give a few hints in order for you to solve the 1.part:

Clearly, due to friction, mechanical energy is not conserved; you must take into account the work done by friction.
To help you on a bit, I'll give the magnitude of the frictional force:
$$F=\mu{N},N=mg\cos(25)$$

5. May 24, 2004

### NonGenius

It's Klassic. A little homework trouble. THANKS SO MUCH.

Did you guys get for number one: 18.08.

As for the second question, I'm a little confused. If you could elaborate, that'd be great. :)

Thanks again!

Last edited: May 24, 2004
6. May 25, 2004

### Klassic

Alright, thanks guys. I definitely figured out #2 - thanks, aril, I figured out how to apply that. Many thanks.

As for #1, I think I got it: 18.08, but I would really like a confirmation if any of you are willing because I'm not entirely content. I used both equations you two gave me.

Again, much appreciated.

Thanks~

7. May 25, 2004

### arildno

Unfortunately, you have round-off errors in your first answer; it should be EXACTLY 18!
Reason:
We have:
$$mg=m\frac{V^{2}}{r}$$
$$\frac{1}{2}mV^{2}=mg\bigtriangleup{h}$$

Eliminating factor mV^{2} from the first equation yields:
$$\frac{mgr}{2}=mg\bigtriangleup{h}\rightarrow\bigtriangleup{h}=\frac{r}{2}$$