# An Olympic skier moving at 18 m/s down a slope encounters a region of wet snow of coefficient of kinetic friction = 0.55. How far down the slope does she travel before coming to a halt? g = 9.8...

An Olympic skier moving at 18 m/s down a slope encounters a region of wet snow of coefficient of kinetic friction = 0.55. How far down the slope does she travel before coming to a halt? g = 9.8 m/s^2

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As the angle made by the slope with the horizontal is not given, it is taken to be A.

The vertical force of gravitational attraction can be divided into two components, one that is parallel to the slope and the other that is perpendicular to the slope.

The component parallel to the slope is M*g*sin A and the component that is perpendicular or normal to the slope is M*g*cos A.

As the coefficient of kinetic friction is 0.55, the frictional force is M*g*cos A*0.55. The deceleration due to this is g*cos A*0.55

The skier is being accelerated down the slope at g*sin A. The net acceleration downwards is g*sin A - g*cos A*0.55.

For the skier to come to a stop g(sin A - 0.55*cos A) should be negative. Then the distance D for the skier to come to a stop is 0 = 18^2 - g(sin A - 0.55*cos A)*D

=> D = 18^2/g(sin A - 0.55*cos A)

The distance traveled by the skier down the slope to come to a stop when the angle made by the slope with the horizontal is A is given by 18^2/g(sin A - 0.55*cos A) m.