# Work done by friction on a skier and resulting distance the skier travels

by Sequence1123
Tags: distance, friction, resulting, skier, travels, work
 P: 2 1. The problem statement, all variables and given/known data A skier starts from rest on a 20m high, 20° slope. μk=0.210 Find the horizontal distance traveled by the skier. From this, for the equations below we know that: yf = 0 vi = 0 vf = 0 2. Relevant equations Wnet = Wnc + Wg = ΔKE Wnet = -fkd Wnc = ΔKE + ΔPE Wnc = ΔKE + mg(yf - yi) KE = 1/2mv2 PE = mgy fk = μkmg 3. The attempt at a solution So I went with the work of a non-conservative force Wnc = (KEf - KEi) + (PEf - PEi) Wnc = (1/2mvf2 - 1/2mvi2) + (mgyf - mgyi) From given, I eliminated all 0 quantities, leaving me with Wnc = -mgyi Then plugged in Wnet = -fkd = -μkmgd so, -μkmgd = -mgyi eliminated like terms (m, g): -μkd = -yi and solved for d d = yi/μk 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
Mentor
P: 40,261
 Quote by Sequence1123 1. The problem statement, all variables and given/known data A skier starts from rest on a 20m high, 20° slope. μk=0.210 Find the horizontal distance traveled by the skier.
Is this the full statement of the problem?

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.
P: 2
 Quote by Doc Al Is this the full statement of the problem? 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.
AH! I figured it out...

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.

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