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

  • Thread starter Thread starter sourna
  • Start date Start date
  • Tags Tags
    Angle
AI Thread Summary
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.
sourna
Messages
2
Reaction score
0

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
 
Physics news on Phys.org
Welcome to Physics Forums.
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.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How to calculate the kinematics of an object pushed up an inclinex plane?
Replies
7
Views
1K
How Do You Calculate Skier's Mass and Acceleration with Newton's Second Law?
Replies
2
Views
1K
Optimized Kinetic Energy of Skier on Incline & Horizontal Surface
Replies
23
Views
2K
Dynamics Skier on a Slope Problem
Replies
9
Views
5K
Terminal velocity of a skier using a Momentum Balance
Replies
1
Views
2K
Solving Work & Forces Problems: Alpine Rescue Team
Replies
1
Views
2K
Speed skier travelling up a hill
Replies
7
Views
3K
A ball (sphere) rotating along a moving incline
Replies
14
Views
2K
How Do Frictional Forces Affect a Skier's Acceleration?
Replies
4
Views
2K
What is the coefficient of friction in this inclined plane problem?
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top