Coefficient of Friction of a olympic skier

In summary: In each equation, you will have to solve for f. 1. In the x direction, you have: Fn=mg+Ff 2. In the y direction, you have: Ff=mg+Fn 1. In the x direction, you have: Fn=mg+Ff 2. In the y direction, you have: Ff=mg+Fn Now, you can solve for f in each equation. 1. In the x direction, you have: Fn=mg+Ff 2. In the y direction, you have: Ff=mg+Fn f=0.871
  • #1
William Bush
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1. The problem statement
An olympic skier moving at 20.0 m/s down a 30.0 degree slope encounters a region of wet snow and slides 145 m before coming to a halt. What is the coefficient of friction between the skis and the snow?




2. Homework Equations
Mk=Ff/Fn
FdCosO=1/2MVf^2-1/2MVo^2




3. The Attempt at a Solution
Drew free body diagram and identified the forces acting on the skier to be the normal force (Fn), gravity (mg), and Ff (frictional force). Fn and mg cancel each other out so the only other force to consider is Ff. I figured I could use the work/kinetic energy formula to find Ff then plug the answer into the formula for Mk. Problem is that I need the mass of the skier for my plan to work.
 
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  • #2
think of the equation that does not require mass. (or it might cancel out!)
 
  • #3
The only equation I can think of where mass will cancel out is: PE (potential energy) at the top equals KE (kinetic energy at the bottom). I don't think this is the equation you are referring to is it?
 
  • #4
think of it this way. The skier skied down with velocity 20 m/s and 30 degrees downwards stopped (or is in rest) in distance 145 meters.
would the mass of the skier be really necessary?
 
  • #5
It's obvious that I missing a key concept. With info given in the problem, I can find acceleration, the height of the slope and even the time it took to travel 145 meters. But I'm not seeing a way to find the frictional force. I am 3 months into my very first physics class so it's possible that there is another formula or concept that I am not remembering.
 
  • #6
hmm

i kept thinking that this problem was to find time!
I believe u need the mass to find the friction force.
 
  • #7
No you don't need the mass, since you will know the deceleration caused by the friction force. And as acceleration is Force per mass, the mass will cancel out each other when you make your calculation later.
 
  • #8
I'm still not following...how do I determine deceleration?
 
  • #9
William Bush said:
I'm still not following...how do I determine deceleration?

You can just use kinematics to get the deceleration, using the information about the velocity and distance stated in the question.

After that, draw a free body diagram and sum up the forces in each direction. That will give you everything you need to solve for the coefficient of friction.
 
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  • #10
1. Is deceleration the same as acceleration but just in the opposite direction?

2. When I use the kinematics equation Vf^2=Vo^2+2ad, I get 1.38m/s for acceleration/deceleration. If I'm right so far, how do I use that info?

3. After drawing the free body diagram, I have three forces; Fn, Ff, and mg. Fn and Ff cancel each other out so the only one I have to be concerned with is Ff. I don't know how to use the value I obtained for deceleration to help me find Ff.
 
  • #11
1. Yes. The word deceleration is sometimes used to indicate that an object is slowing down (rate of change of velocity is decreasing).

2. That's right, but watch your signs. You're going to be using Newton's second law, so that's why you need to know acceleration.

3. I don't understand this:

I have three forces; Fn, Ff, and mg. Fn and Ff cancel each other out so the only one I have to be concerned with is Ff.

Fn and Ff cancel, but you're left with Fn?? What I think you meant to say (correct me if I'm wrong) is that Fn and mg cancel. Well, remember, the skier is on an incline. Have you learned how to break up forces into x and y components? That is what you need to do here. You will use Newton's second law to get two equations. One in the x direction, and one in the y direction.
 

FAQ: Coefficient of Friction of a olympic skier

What is the coefficient of friction of an Olympic skier?

The coefficient of friction of an Olympic skier refers to the measure of the amount of friction that exists between the skier's equipment and the snow surface. It is a dimensionless quantity that is determined by the type of skis, wax, and snow conditions.

Why is the coefficient of friction important for Olympic skiers?

The coefficient of friction is important for Olympic skiers because it affects their speed and control on the slopes. A higher coefficient of friction means more resistance, resulting in slower speeds, while a lower coefficient of friction allows for faster speeds and better maneuverability.

How is the coefficient of friction of an Olympic skier measured?

The coefficient of friction of an Olympic skier is typically measured in a controlled laboratory setting using a device called a tribometer. This device measures the frictional force between two surfaces, such as the ski and the snow, and calculates the coefficient of friction.

What factors affect the coefficient of friction of an Olympic skier?

The coefficient of friction of an Olympic skier can be influenced by various factors, including the type of skis, wax, and snow conditions. Additionally, the skier's body weight, speed, and technique can also affect the coefficient of friction.

Can the coefficient of friction of an Olympic skier be changed?

Yes, the coefficient of friction of an Olympic skier can be altered by changing the type of skis, wax, or snow conditions. Skiers can also adjust their technique to reduce or increase the amount of friction and improve their performance on the slopes.

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