A62.0-kg skier is moving at 6.50 m/s on a frictionlesshorizontal snow-covered plateau when she encounters a rough patch 3.50 mlong. the coefficient of kinetic friction between this patch and her skisis 0.300. after crossing the rough patch and returning to friction-free snow, she skis downan icy, frictionless hill 2.50 m high.(a) how fast is the skier moving when she gets to the bottom ofthe hill? (b) how much internal energy was generated in crossing the roughpatch?

(A) 5.22 m/s      

(B) 637.98 j

Explanation:

from the question we are given the following

mass of the skier (m) = 62 kg

initial velocity of the skier (U1) =6.5 m/s

length of the rough path (s) = 3.5 m

coefficient of kinetic friction () = 0.3

height of the hill (y) = 2.5 m

acceleration due to gravity (g) = 9.8 m/

final velocity (v) =?

(a) To solve this problem we would split the motion of the skier into two phases:

phase 1 would be from when he starts to when he gets to the icy slope

phase 2 would be down the icy slope

we would be applying the equation below

work done due to gravity + work done due to friction = K2 - K1

where K1 and K2 are the kinetic energies

kinetic energy =

work done due to gravity = m x g x height of the slope

work done due friction = x m x g x s

Phase 1:

when the skier moves across the rough patch the the work done due to gravity is 0 since he moves horizontally and perpendicular to the gravitational force. Therefore the equation now becomes

work done due to friction = K2 - K1

-(0.3 x 62 x 9.8 x 3.5 =  ^{2}" alt="U_{2}" />^{2}" />) -

-637.98 = ^{2}" alt="U_{2}" />^{2}" /> - 1309.75

U_{2} = 4.66 m/s

Phase 2

when the skier moves down the hill on a friction free snow, work is only done by gravity. therefore the equation becomes

work done due to gravity = K3 - K2

62 x 9.8 x 2.5 = ^{2}" alt="U_{3}" />^{2}" />) -

1519 = 31 ^{2} + 673.18

U_{3} = 5.22 m/s      

at the end of the slope the skier's velocity is 5.22 m/s      

(B)  the internal energy generated while crossing the rough path is the same as the work done due to friction = 637.98 j

(this is because internal energy is the manifestation of work done by non-conservative forces, and frictional force is a non-conservative force).

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq