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A car is driving down a mountain ($v=90 km/h=25 m/s$, when the driver realizes that brakes aren't working. He try to lose velocity going up an inclined ($20°$) plane, with a friction coefficient of $k=0.60$. How many meters will it take to halt?

I've tried as following ($s$ is the request): $$K=\frac{mv^2}{2}$$

At the end, the potential energy gained is: $$U=mgh=mg\cdot s\cdot sin \alpha$$

In the mainwhile the energy lost due to the friction is: $$L_f=F \cdot s=mg \cdot cos(\alpha) \cdot s$$

But the work done by non conservative forces (friction) is also: $$L_f=U-K$$

And I have: $$mg \cdot cos(\alpha) \cdot s=mg\cdot s\cdot sin \alpha-\frac{mv^2}{2}$$ $$g \cdot cos(\alpha) \cdot s=g\cdot s\cdot sin \alpha-\frac{v^2}{2}$$ $$9.22s=3.35s-312.5$$ But I get a negative time. What's wrong? I'm sure that there is a stupid error, but I can't find it.

The correct result (reported on the textbook) is 120 m.

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Shouldn’t it be $K = U + L_f$, as the initial kinetic energy is equal to the potential energy at the end plus the energy lost due to friction? –  Claudius Dec 5 '12 at 16:26

1 Answer 1

up vote 1 down vote accepted

There are two sources of kinetic energy loss. a) is friction and the other b) conversion to potential energy. So

$$ \frac{1}{2} m v^2 = m g \left( k \cos\alpha + \sin\alpha \right) s $$

In your post you have as if gravity is providing energy to the system which it would if the slope was downwards. If you correct your sign you will get the correct result.

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The correct result (reported on the textbook) is 120 m. If I do what you say, the result is 35,17 m, isn't it? Thanks –  Surfer on the fall Dec 5 '12 at 16:40
    
Your textbook is wrong then! –  Jaime Dec 5 '12 at 18:46

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