A 62.0 kg skier is moving at 6.80 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.80 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to the friction-free snow, she skis down an icy, frictionless hill 2.50 m high.(a) How fast is the skier moving when she gets to the bottom of the hill?(b) How much internal energy was generated in crossing the rough patch?

Explanation:

force of friction = μ mg

= .3 x 62 x 9.8

= 182.28 N

Work done by friction = 182.28 x 4.8

= 874.94 J

Energy loss by skier = 874.94 J

Kinetic energy after he passes the rough patch

= 1/2 x 62 x 6.8² - 874.94

= 1433.44 - 874.94

= 558.5 J

Velocity after crossing the patch v

1/2 x 62 x v² = 558.5

v² = 18

v = 4.2426 m /s

velocity at the bottom

v² = u² + 2gh

= 18 + 2 x  9.8 x 2.5

= 67

v = 8.18 m /s

b ) internal energy generated =

Work done by friction  . = 874.94 J

the part of the wave he is travelling on is the crest