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Find the coefficient of kinetic friction

by yanase
Tags: coefficient, friction, kinetic
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yanase
#1
May9-12, 03:30 AM
P: 2
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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azizlwl
#2
May9-12, 04:32 AM
P: 963
Apply work-energy principle.
Doc Al
#3
May9-12, 08:40 AM
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P: 41,323
Your work looks OK to me.
Quote Quote by yanase View Post
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.)

azizlwl
#4
May9-12, 08:30 PM
P: 963
Find the coefficient of kinetic friction

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
azizlwl
#5
May9-12, 09:01 PM
P: 963
Quote Quote by yanase View Post

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.
Doc Al
#6
May10-12, 06:10 AM
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P: 41,323
Quote Quote by azizlwl View Post
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
#7
May10-12, 06:19 AM
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P: 41,323
Quote Quote by yanase View Post
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!
azizlwl
#8
May10-12, 09:22 AM
P: 963
Thank Doc Al, I'm too looking for Yanase's error which lead to wrong answer and your advice of using multiple methods.
yanase
#9
May12-12, 12:41 AM
P: 2
Thank you very much to both of you. I really appreciate you help in solving my question. :D


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