text: # Homework Help: Finding Speed and Coefficient of Friction
  
  1. Oct 2, 2004
  
  Finding Speed with Coefficient of Friction
  
  You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-27). You find that the slope of the hill is = 12.0°, that the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.5 m/s.
  
  With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
  
  I've figured out this equation since there is constant acceleration:
  V^2=Vo^2+2a(x-xo) where a=-(Mk)g. I haven't taken into account the incline and I'm not sure how. How do I incorporate the incline into this equation, if my equation is even right.
  
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  Last edited: Oct 2, 2004
  2. Oct 2, 2004
  
  ### Pyrrhus
  
  Let me point out change of Mechanical Energy will be equal to the work done by friction.
  
  $$\Delta E = W_{f}$$
  
  Last edited: Oct 2, 2004
  3. Oct 2, 2004
  
  I don't see the relation....
  
  4. Oct 2, 2004
  
  ### Pyrrhus
  
  $$E - E_{o} = W_{f}$$
  
  What's not to see?
  
  Tell me exactly what you don't understand.
  
  5. Oct 2, 2004
  
  Well in my class we haven't dealt with that equation yet, and this problem should have nothing to do with Mechanical Energy, just Newton's Laws and Acceleration and Friction.
  
  6. Oct 2, 2004
  
  ### Pyrrhus
  
  Oh, should had said so , then use newton's 2nd law (to find acceleration) and kinematics for final speed.
  
  7. Oct 2, 2004
  
  I can't do that lol I wrote out my equation in my original post I just can't incorporate the 12 degree angle.
  
  8. Oct 2, 2004
  
  ### Pyrrhus
  
  Put your X-axis along the 12 degree straight line, so the normal force will have only a non-zero component, and the weight of the car will have 2 non-zero components one pointing left, and the other pointing down., and the friction force will be on the right, against movement with 1 non-zero component.
  
  Last edited: Oct 2, 2004
  9. Oct 2, 2004
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