Finding Speed and Coefficient of Friction

[GingerBread27]

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.

[Pyrrhus]

Let me point out change of Mechanical Energy will be equal to the work done by friction.

\Delta E = W_{f}

[GingerBread27]

I don't see the relation....

[Pyrrhus]

E - E_{o} = W_{f}

What's not to see?

Tell me exactly what you don't understand.

[GingerBread27]

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.

[Pyrrhus]

Oh, should had said so :smile: , then use newton's 2nd law (to find acceleration) and kinematics for final speed.

[GingerBread27]

I can't do that lol I wrote out my equation in my original post I just can't incorporate the 12 degree angle.

[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.

[GingerBread27]

hmm not understanding

[Pyrrhus]

Draw a Free body diagram.