# Find the coefficient of kinetic friction

## 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.

Apply work-energy principle.

Last edited:
Doc Al
Mentor
Your work looks OK to me.
g(sin θ - μcos θ)= -(μgsB)/sA
μ(cos θ - sA/sB) = sin θ
μ = (sin θ)/(cos θ - sA/sB)
I think you have a typo here. That ratio should be sB/sA, not sA/sB.

Substituting θ=17degree, sA=100m, sB=70m into the equation,
μ = (sin 17)/(cos 17 – 70/100)
= 1.14
But it looks like you plugged in the correct ratios.

The value of μ should be between 0 and 1. Can anyone tell me where I went wrong in solving this question?
I don't think you went wrong. The data is just unrealistic.

Edit: I found the mistake; see my post #7. (And please look at azizlwl's alternative solution. It's always good to solve things multiple ways.)

Last edited:
ΔPE+ΔKE=Wf
PEi=mgh=f(x_slope)+f(x_level)
PEf=0,KEi=KEf=0
m=90kg
h=100sin17°
length on slope=100m
length on level =70m

Mg(100sin17°)=μN(x_slope)+μN(x_level)
Mg(100sin17°)=μMgCos17°(100) +μMg(70)
μ=100sin17°/(100Cos17° +70)
μ=29.23/165.63
μ=0.17

B:
-Ff = maB ---------------- 6
down the hill
N - mg = 0 → N = mg ---- 7

3→6
-μN = maB --------------- 8

6→7
-μmg = maB
aB = -μg

Deleting my comment on this.

Last edited:
Doc Al
Mentor
ΔPE+ΔKE=Wf
PEi=mgh=f(x_slope)+f(x_level)
PEf=0,KEi=KEf=0
m=90kg
h=100sin17°
length on slope=100m
length on level =70m

Mg(100sin17°)=μN(x_slope)+μN(x_level)
Mg(100sin17°)=μMgCos17°(100) +μMg(70)
μ=100sin17°/(100Cos17° +70)
μ=29.23/165.63
μ=0.17
Looks good to me.

I just redid the problem using the kinematic method used in post # 1 and got the same answer.

I'll have to look to see where the error in post #1 is. Oops!

(I see the mistake... a sign error!)

Doc Al
Mentor
using v2=u2+2as ;
u=0 since it starts from rest,
v2=2as

vA2=2aAsA
vB2=2aBsB
Careful here. For part A, the speed goes from 0 to V; but for part B the speed goes from V to 0.

So: 0 - VB2 = 2aBsB.

You have a sign error in applying the kinematic formula to the horizontal motion. Fix that and you'll get a sensible answer.

Sorry for not spotting that earlier.

Thanks to azizlwl!