# Not working brakes: just another energy conservation problem

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

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