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posted by  jojo5 on 7/4/2009 1:28:00 AM  |  status: Live  |  Earned Karma: 25

Physics

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Question Details:
A box is sliding up an incline that makes an angle of 15 degrees with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is .180. The initial speed of the box at the bottom of the incline is 1.50 m/s. How far does the box travel along the incline before coming to rest?
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posted by jojo5 on 7/4/2009 2:21:14 AM  |  status: Live
Clarification Details:
The book says that the answer is 0.265m. How do you get this answer?

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posted by watt (MNK) on 7/4/2009 1:35:03 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:
angle θ = 15 degrees
The coefficient of kinetic friction between the box and the surface of the incline  μ = 0 .180.
The initial speed of the box at the bottom of the incline  v = 1.50 m/s
Final speed  V = 0 m/ s
Accleration a = -μg sin θ
                    = - 0.4565 m / s ^ 2
from the relation V ^ 2 - v ^ 2 = 2aS 
distance moved  S = 2.46 m
(SME)
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posted by Kevin_08 on 7/4/2009 1:39:14 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:
  From the above free body diagram
         F =f
             = μ mg sinθ
  From the work -energy theorm
          Fs = (1/2)mv2 - (1/2)mu2
                   
  Here v=0  and u = 1.50 m/s
           ( μ mg sinθ) s =  (1/2)mv2 - (1/2)mu2
  The distance travelled by the box is
            s = [1/2)v2 - (1/2)u2]/( μ g sinθ)
 Here θ = 15 degrees
           g =9.8 m/s2
        μ = 0.18
Substitute the values for s.
I hope this helps! Best of luck with the rest of your coursework.
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