6.27: a) The friction force is mg k μ , which is directed against the car’s motion, so the net work done is mgs k-. The change in kinetic energy is 20 1 ) 2 / 1 ( mv K K-=-= ∆ , and so g v s k 20 2 / = . b) From the result of part (a), the stopping distance is proportional to the
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