Coefficents of Friction

  • Thread starter sonoftunk
  • Start date
  • #1
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Homework Statement


A young skier (25kg) pushes off with ski poles to give herself an initial velocity of 3.5 m/s down a hill with 5o slope with a coefficient of friction of 0.20.
Find the time until skier comes to stop and her displacement.
Therefore
m = 25kg
vi = 3.5m/s
[tex]\theta[/tex] = 5o
[tex]\mu[/tex] = 0.20

Homework Equations


All kinematic-based equations
Newton's Laws (specifically [tex]\stackrel{\rightarrow}{F}[/tex] = m([tex]\Delta[/tex][tex]\stackrel{\rightarrow}{v}/[/tex][tex]\Delta[/tex]t)
FF = [tex]\mu[/tex]FN


The Attempt at a Solution


My idea for this was to find the FNET and sub that into Newton's Second Law, solving for [tex]\Delta[/tex]t (assuming that [tex]\stackrel{\rightarrow}{v}[/tex]f is zero)
My solution for [tex]\Delta[/tex]t was 1.3 seconds, correct to two significant digits (to me, that seems quite off)

For the second half of the question, finding the distance, I used [tex]\Delta[/tex][tex]\stackrel{\rightarrow}{d}[/tex] = [tex]\stackrel{\rightarrow}{v}[/tex]i + [tex]\frac{1}{2}[/tex][tex]\stackrel{\rightarrow}{a}[/tex]([tex]\Delta[/tex]t)2 using [tex]\stackrel{\rightarrow}{a}[/tex] as the component force of gravity acting parallel to the hill and I got 23m, correct to 2 significant digits.

I was then told I completed the question completely wrong. Now I'm lost. What was I supposed to do?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
5
Welcome to PF.

Maybe consider it is a Kinetic Energy to work from friction problem?

Work would be the (μ*mg*cos5° - mg*sin5°) times the distance and that would equal the KE.
 
  • #3
5
0
I haven't learned kinetic energy yet...

BTW, this is a correspondence coarse for Grade 12 Physics in Ontario..
 
  • #4
532
2
You know the frictional force, so find the net acceleration on the skier due to friction and gravity.
 

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