A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.

How fast is the skier moving when she gets to the bottom ofthe hill?

Explanation:

The given data is as follows.

      Mass, m = 62 kg,       Initial speed, = 6.90 m/s

 Length of rough patch, L = 4.50 m,      coefficient of friction, = 0.3

 Height of inclined plane, h = 2.50 m

According to energy conservation equation,

     (Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

Cancelling the common terms in the above equation, we get

                         =

                         = 36.055 - 13.23

                         = 22.825

                           = 6.75 m/s

Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.

neither accurate nor precise.