Skier traveling uphill with and without friction

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Homework Help Overview

The discussion revolves around a physics problem involving a skier traveling uphill on a slope with and without friction. The original poster presents a scenario where the coefficient of kinetic friction is given, and the slope is at 30 degrees, seeking to determine how far the skier travels up the hill.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the forces acting on the skier, including friction and gravity, and their effects on the skier's motion. There are attempts to clarify the components of gravitational force on an incline and how these relate to the skier's deceleration.

Discussion Status

The discussion includes various attempts to analyze the forces at play and their contributions to the skier's motion. Participants are actively questioning assumptions and definitions, particularly regarding the effects of slope inclination on gravitational force. Some guidance has been provided regarding the relationship between friction, normal force, and the components of gravitational force.

Contextual Notes

Participants are working within the constraints of a homework problem, which includes specific values for the coefficient of friction and slope angle. There is an ongoing exploration of how these values influence the skier's distance traveled up the hill.

  • #31
Then the total force decelerating the skier is ... ?
 
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  • #32
Bystander said:
Then the total force decelerating the skier is ... ?
mg*cos30 + mg*sin30 ??!??
 
  • #33
You forgot the coefficient of friction.
 
  • #34
Bystander said:
You forgot the coefficient of friction.
OHHH!
μ*mg*sin30 + mg*cos30
 
  • #35
Close. Frictional force is the product of coefficient of friction and the normal force. One more time.
 
  • #36
Bystander said:
Close. Frictional force is the product of coefficient of friction and the normal force. One more time.
okay so if Ff = μ*Fn then...
μ*mg*cos30 + mg*sin30
 
  • #37
Correct. And you've done the d = (1/2)at2 game for the other part of the problem once already, so you should be in business. "Windoze" insists it's going to update, so I'll be off for the next half hour.
 
  • #38
Bystander said:
Correct. And you've done the d = (1/2)at2 game for the other part of the problem once already, so you should be in business. "Windoze" insists it's going to update, so I'll be off for the next half hour.
thank you so much! you have no idea how much that helped me
 

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