A 62.0-kg skier is moving at 6.80 m/s on a frictionless, horizontal, snow-covere

Question

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 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 of the hill?

Express your answer with the appropriate units.

How much internal energy was generated in crossing the rough patch?

Express your answer with the appropriate units.

Explanation / Answer

Here ,

mass of skier , m = 62 Kg

initial speed , u = 6.8 m/s

distance d, = 4.8 m

uk = 0.30

height , h = 2.50 m

let the speed of skier when she get to the bottom is v

Using conservation of energy

0.5 * m * v^2 = 0.5 * m * u^2 + m *g * h - uk * m *g * d

0.5 * 62 * v^2 =0.5 * 62 * 6.8^2 + 62 * 9.8 * 2.50 - 0.30 * 62 * 9.8 * 4.80

solving for v

v = 8.19 m/s

the speed of the skier at the bottom is 8.19 m/s

Internal energy generated in crossing the rough patch = work done by friction

Internal energy generated in crossing the rough patch = 0.30 * 62 * 9.8 * 4.80

Internal energy generated in crossing the rough patch = 875 J

the Internal energy generated in crossing the rough patch is 875 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.