- #1
learningisfun
- 22
- 0
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.
Last edited: