Physics, 11.11.2019 19:31, yeeeeeeeeeeet21

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?

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answered: book0001

(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 ( $u_{k}$ ) = 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 = $\frac{1}{2} mv^{2}$

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 = $\frac{1}{2} 62 x <img src=$ ^{2}" alt="U_{2}" />^{2}" />) - $\frac{1}{2} 62 x 6.5^{2}$

-637.98 = 31<img src= ^{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 = $\frac{1}{2} 62 x <img src=$ ^{2}" alt="U_{3}" />^{2}" />) - $\frac{1}{2} 62 x 4.66^{2}$

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).

answered: Guest

Neptune is the eighth planet from the sun. it was the first planet to get its existence predicted by mathematical calculations before it was actually seen through a telescope on sept. 23, 1846. irregularities in the orbit of uranus led french astronomer alexis bouvard to suggest that the gravitational pull from another celestial body might be responsible. german astronomer johann galle then relied on subsequent calculations to spot neptune via telescope. previously, astronomer galileo galilei sketched the planet, but he mistook it for a star due to its slow motion. in accordance with all the other planets seen in the sky, this new world was given a name from greek and roman mythology — neptune, the roman god of the sea.

answered: Guest

Ultra-high temperature (uht)

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A62.0-kg skier is moving at 6.50 m/s on a frictionlesshorizontal snow-covered plateau when she encou...