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Skier skies down slope, including friction, calculate final speed

  • Thread starter wiggle
  • Start date
1. The problem statement, all variables and given/known data

A 50Kg skier skies down a 25 degree slope. at the top of the slope her speed is 4m/s and accelerates down the hill. the coefficient of friction is 0.12 between skies and snow. ignoring air resistance calculate her speed at point that is displaced 60m downhill.


2. Relevant equations

E=KE+PE
E=.5*m*V^2+m*g*h


3. The attempt at a solution

I believe I solved it correctly using E0=.5(50)4^2+50(9.8)(60sin(25)) and getting 12650

then setting 12650 = .5*m*Vf^2 and I'm getting Vf=22.5 m/s

However I'm unsure how to go about solving while taking into consideration the friction, since there is a non conservative force at work using KE0+PE0=KEf+PEf would get me close but not all the way there.
 

Doc Al

Mentor
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The work done by friction will reduce the total mechanical energy.
 
Right.

So I've just broken it up into x and y components

Fx=Wsin25-.12(Wcos(25))=m*a

I calculated for acceleration in the x plane, and found a=3.08 m/s^2

Then I just did

Vf=sqr(2ax+V0^2)
sqr(2(3.08)(60)+4^2)= Vf= 19.6 m/s

Does this seem correct?
 

Doc Al

Mentor
44,725
1,021
Looks good to me.
 

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