- #1

- 2

- 0

## Homework Statement

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?

## Homework Equations

θ=17degree, sA=100m, sB=70m, m=90kg

## 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.