How Do You Calculate the Coefficient of Friction for a Skier on an Incline?

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To calculate the average coefficient of friction for a skier on a 20° incline, the skier's initial speed of 12.6 m/s and the distance traveled of 11.4 m must be considered. The energy conservation equation Egi + Eki + wnc = Egf + Ekf is relevant, where wnc relates to friction. Clarification was provided that the 11.4 m represents the distance along the slope, not the vertical height. Understanding this distinction is crucial for accurately determining the gravitational energies involved. Proper application of these concepts will lead to the calculation of the coefficient of friction.
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Homework Statement


A skier traveling 12.6 m/s reaches the foot of a steady upward 20° incline and glides 11.4 m up along this slope before coming to rest. What was the average coefficient of friction?

Homework Equations


Egi+Eki+wnc=Egf+Ekf

The Attempt at a Solution


I solved for everything but i don't know how to solve for mew
 
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sourna said:

Homework Statement


A skier traveling 12.6 m/s reaches the foot of a steady upward 20° incline and glides 11.4 m up along this slope before coming to rest. What was the average coefficient of friction?

Homework Equations


Egi+Eki+wnc=Egf+Ekf

The Attempt at a Solution


I solved for everything but i don't know how to solve for mew
Can you show what you did with the equation by using the given information? Also, you'll need to think about how wnc is related to friction.
 
Redbelly98 said:
Welcome to Physics Forums.

Can you show what you did with the equation by using the given information? Also, you'll need to think about how wnc is related to friction.
is it ok if i say no? I don't understand it very well you see because I am not sure if 11.4 m is height or distance. :confused:
 
Okay. 11.4 m is the distance along the slope, not the height. That should help you with finding Egi and Egf in the equation you wrote earlier.
 

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