(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A skier weighing 90kg starts from rest down a hill inclined at 17 degree. He skis down the hill and then coast for 70 m along level snow until he stops. Find the coefficient of kinetic friction between the skis and the snow. What velocity does the skier have at the bottom of the hill?

2. Relevant equations

θ=17degree, sA=100m, sB=70m, m=90kg

3. The attempt at a solution

A-motion skis down the hill

B-motion along level snow

A:

mg sin θ - Ff = maA ------ 1 ;Ff is the frictional force

N - mg cos θ = 0

→ N = mg cos θ -------- 2

Ff = μN ----------------- 3 ; μ is coefficient of kinetic friction

3→1

mg sin θ - μN = maA ------- 4

2→4

mg sin θ - μmg cos θ = maA

aA = g(sin θ - μcos θ) ---- 5

B:

-Ff = maB ---------------- 6

down the hill

N - mg = 0 → N = mg ---- 7

3→6

-μN = maB --------------- 8

6→7

-μmg = maB

aB = -μg

using v2=u2+2as ;

u=0 since it starts from rest,

v2=2as

vA2=2aAsA

vB2=2aBsB

vA2 = vB2

2aAsA = 2aBsB

aA = -(μgsB)/sA ; aB = -μg

g(sin θ - μcos θ)= -(μgsB)/sA

μ(cos θ - sA/sB) = sin θ

μ = (sin θ)/(cos θ - sA/sB)

Substituting θ=17degree, sA=100m, sB=70m into the equation,

μ = (sin 17)/(cos 17 – 70/100)

= 1.14

The value of μ should be between 0 and 1. Can anyone tell me where I went wrong in solving this question? Thank you.

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# Find the coefficient of kinetic friction

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