- #1
Oliviam12
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The Question is:
A skier goes down a slope with an angle of 35 degrees relative to the horizontal. Her
mass, including all equipment, is 70 kg. The coefficient of kinetic friction between her
skies and the snow is 0.15.
A. Please draw a free-body diagram of the skier.
B. Calculate the net force acting on the skier.
C. If the slope is 60 m long, what is her speed at the bottom of the slope,
assuming that she started from rest?
Does this like correct? (Especially B and C, seeing as I have never done this type before)
My A is:
http://img88.imageshack.us/img88/1207/freend1.th.png
My B is:
Fg=-mg
Fg=-70(9.81)
Fg=-686.7 N
Fk= [tex]\mu[/tex]Fg
Fk= .15 (-686)
Fk = -102.9 N
Net Force: -686.7 - -102.9= -583.8 NC.) (The length of the ramp dosn't really matter does it?)
F=MA
-583.8= 70A
-8.34 m/s^2 =A
Thanks!
A skier goes down a slope with an angle of 35 degrees relative to the horizontal. Her
mass, including all equipment, is 70 kg. The coefficient of kinetic friction between her
skies and the snow is 0.15.
A. Please draw a free-body diagram of the skier.
B. Calculate the net force acting on the skier.
C. If the slope is 60 m long, what is her speed at the bottom of the slope,
assuming that she started from rest?
Does this like correct? (Especially B and C, seeing as I have never done this type before)
My A is:
http://img88.imageshack.us/img88/1207/freend1.th.png
My B is:
Fg=-mg
Fg=-70(9.81)
Fg=-686.7 N
Fk= [tex]\mu[/tex]Fg
Fk= .15 (-686)
Fk = -102.9 N
Net Force: -686.7 - -102.9= -583.8 NC.) (The length of the ramp dosn't really matter does it?)
F=MA
-583.8= 70A
-8.34 m/s^2 =A
Thanks!
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