2- Three sisters go skiing. Each has a mass of m = 50 kg. There are three differ

Question

2- Three sisters go skiing. Each has a mass of m = 50 kg. There are three different routes to the bottom. Route A: a constant slope of 3.81 degrees for 12,027 meters. Route B: a gentle slope of 1.83 degrees for 25,013 meters. Route C: a steep slope of 18.43 degrees for 2527 meters. Each sister takes one route, and each starts from rest at the top of the hill letting gravity do the work. Ignore friction.
(a) How much work does gravity do on each sister during the descent?
(b) How fast is each sister going when she reaches the bottom?
The sisters ski down the same routes as before a second time. Now, however, the snow has melted slightly and become sticky. The coefficient of kinetic friction is µk = 0.02.
(c) How much work does friction do on each sister during the descent?
(d) Now how fast is each sister going when she reaches the bottom?

Explanation / Answer

a.) Work Done By gravity = mgh sin

a) 1st sister = 50*9.8*12027*sin3.81 = 391.6 kJ

2nd   sister = 50*9.8*25013*sin1.83 = 391.4 kJ

3rd   sister = 50*9.8*2527*sin18.43 = 391.5 kJ

b)

Now potential energy converts to ke

==>

mgh sin = 0.5 m v^2

v = (gh sin )/(.5 )

1st sister = 391.6 * 10^3 = 0.5 m v^2

v =  ( 391.6 * 10^3)/(.5 * 50) = 125.1559 m/sec

2nd sister = 391.4 * 10^3 = 0.5 m v^2

v =  ( 391.4 * 10^3)/(.5 * 50) = 125.1239 m/sec

3rd sister = 391.5 * 10^3 = 0.5 m v^2

v =  ( 391.5 * 10^3)/(.5 * 50) = 125.1399 m/sec

C)

uk = 0.02

Work done by friction = mgh cos *uk

1st sister = 50*9.8*12027*cos3.81*0.02 = 117.6 kJ

2nd   sister = 50*9.8*25013*cos1.83*0.02 = 245 kJ

3rd   sister = 50*9.8*2527*cos18.43*0.02 = 234.9 kJ

d)

Now

Potential energy = Kinetic energy + work done by friction

1st sister;

391.6 Kj = .5 mv^2 + 117.6 Kj

solving we get;

v = 104.69 m/sec

2nd   sister

391.4Kj = .5 mv^2 + 245 Kj

solving we get;

v = 76.52 m/sec

3rd   sister

391.5 kJ =  .5 mv^2 + 234.9 Kj

solving we get;

v = 79.14 m/sec

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