Question

A 62.0-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?

Answer 1

(A) 5.22 m/s

(B) 637.98 j

Step-by-step 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 (
`u_(k)`) = 0.3

height of the hill (y) = 2.5 m

acceleration due to gravity (g) = 9.8 m/
`s^(2)`

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 =
`(1)/(2) mv^(2)`

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

work done due friction =
`u_(k)` 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 =
`(1)/(2) 62 x [tex]U_(2)`^{2}[/tex]) -
`(1)/(2) 62 x 6.5^(2)`

-637.98 =
`31[tex]U_(2)`^{2}[/tex] - 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 =
`(1)/(2) 62 x [tex]U_(3)`^{2}[/tex]) -
`(1)/(2) 62 x 4.66^(2)`

1519 = 31
`U_(3)`^{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).