Register to reply 
Work done by friction on a skier and resulting distance the skier travels 
Share this thread: 
#1
Feb2912, 04:43 PM

P: 2

1. The problem statement, all variables and given/known data
A skier starts from rest on a 20m high, 20° slope. μ_{k}=0.210 Find the horizontal distance traveled by the skier. From this, for the equations below we know that: y_{f} = 0 v_{i} = 0 v_{f} = 0 2. Relevant equations W_{net} = W_{nc} + W_{g} = ΔKE W_{net} = f_{k}d W_{nc} = ΔKE + ΔPE W_{nc} = ΔKE + mg(y_{f}  y_{i}) KE = 1/2mv^{2} PE = mgy f_{k} = μ_{k}mg 3. The attempt at a solution So I went with the work of a nonconservative force W_{nc} = (KE_{f}  KE_{i}) + (PE_{f}  PE_{i}) W_{nc} = (1/2mv_{f}^{2}  1/2mv_{i}^{2}) + (mgy_{f}  mgy_{i}) From given, I eliminated all 0 quantities, leaving me with W_{nc} = mgy_{i} Then plugged in W_{net} = f_{k}d = μ_{k}mgd so, μ_{k}mgd = mgy_{i} eliminated like terms (m, g): μ_{k}d = y_{i} and solved for d d = y_{i}/μ_{k} and plugged in the knowns d = 20m/0.210 d = 95.2m I realize this is the distance traveled from the top of the hill to the end of motion, but all they want is the horizontal distance, so now I have to solve for the distance traveled from the top of the hill to the bottom of the hill. The only thing I think that changes between the above work and the distance from the top to bottom is the final velocity which will be nonzero. So my question is, how do I find the distance traveled from the top of the hill to the bottom? Or am I going about this wrong? Is there a more direct way to find just the horizontal distance traveled? Oh I should add, the answer given by the book is 40.3m 


#2
Feb2912, 05:00 PM

Mentor
P: 41,435

I'm guessing that after the skier gets to the bottom of the slope, she skis over a horizontal stretch of ground? Perhaps it's that horizontal distance that they want. 


#3
Feb2912, 05:21 PM

P: 2

I had to find the length of the bottom of the triangle formed by the horizontal and the slope of the hill. sin20° = 20/x (where x is the hypotenuse, or the length of the slope of the hill) from that I got x ≈ 58.5m Then, cos20° = x/58.5m (where x is the length of the bottom of the triangle) x ≈ 54.9m then subtract that from the total length traveled by the skier, 95.2m  54.9m = 40.3m which is the answer given by the book. 


Register to reply 
Related Discussions  
Forces  distance of a skier on an incline  Introductory Physics Homework  3  
Skier on a Slope  Find Total Distance  Introductory Physics Homework  2  
Friction and a skier  Introductory Physics Homework  3  
Need help finding distance of skier going down a hill! Help!  Introductory Physics Homework  6  
Skier velocity and friction problem  Introductory Physics Homework  3 