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## Homework Statement

A sight seen on many bunny hills across ontatrio is young skiers pushing on ski poles and gliding down a slope until they come to a rest . Obeserving from a distance , you note a young person (approximatly 25kg) pushing off with the ski poles to give herself an intial velocity of 3.5m/s. If the inclination of the hills is 5 degress and hte coefficient of kinetic friction for the skis on dry snow is 0.20 calculate.

A) the time taken for the skier to come to a stop

## Homework Equations

solving for t

a=v/t

solving for a

a=Fnet/m

## The Attempt at a Solution

vi-3.5m/s m=25kg mu=0.20 angle=5degress

I'm assuming that the skier without velocity would be stationary on the inclination.

This is my first problem, is she accelerating down the ramp or moving at a constant velocity

Fg =mg=25kg(9.8m/s^2)=245N

Fn=Fg-Fy=mg-mgsin5=25kg(9.8m/s^2)-25kg(9.8m/s^2)sine5=224N

Fg=245N

Fn=224NFf=muFn=0.20mgcos5=49N

49 Newtons on the yaxis

This is where I'm confused

I have another equation that goes like this

Fa=Fgsine5 + muFn

=masine5+mumascos5

=ma(sine +mucos5)

-25kg(9.8m/s^2)(sine5+0.20(cos5))

=70N

I'm assuming this equation accounts for acceleration, subtracting the Ff.

Fnet=x+y=21+70

=91N

a=91/25=3.64m/s^2

t=3.5m/s/3.64m/s^2=0.9s

This is wrongoh and can someoen give me a more detailed explaantion what Fnormal is. I"m assuming it is the x component Fg subtracted from Fg.

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