Skier traveling uphill with and without friction

In summary: Correct. The component perpendicular to the slope is mg*sin(30).Correct. The component perpendicular to the slope is mg*sin(30).In summary, the problem asks for the distance up the hill a skier travels given a coefficient of kinetic friction of 0.11 and a slope of 30 degrees. The solution involves considering the forces of friction and gravity and their effects on the skier's motion. Using trigonometry, the gravitational force can be broken
  • #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
 
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  • #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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